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Unit 1: Limits and Continuity

1. Topic 1.1 · Work it outFind the average rate of change of $f(x)=x^2$ on $[\mathrm{1,3}]$. Then find it on $[\mathrm{1,1}+h]$ and say what happens as $h$ shrinks toward 0.

   On $[\mathrm{1,3}]$: $\frac{9-1}{2}=4$. On $[\mathrm{1,1}+h]$: $\frac{(1+h{)}^2-1}{h}=2+h$, which approaches 2. That limit is the instantaneous rate of change at $x=1$.

   Review: 1.1 Introducing Calculus: Can Change Occur at an Instant?

2. Topic 1.2 · Know the termWhat does ${\lim }_{x\to a}f(x)=L$ say, and does it depend on $f(a)$?

   As $x$ gets arbitrarily close to $a$ (from both sides, $x\ne a$), $f(x)$ gets arbitrarily close to $L$. It does not depend on $f(a)$, which can differ from $L$ or not exist.

   Review: 1.2 Defining Limits and Using Limit Notation

3. Topic 1.2 · Spot the mistakeA student says ${\lim }_{x\to 2}\frac{x^2-4}{x-2}$ does not exist because the function is not defined at $x=2$. What is wrong?

   A limit describes what happens near 2, not at 2. For $x\ne 2$ the expression equals $x+2$, so the values approach 4 from both sides and the limit is 4.

   Review: 1.2 Defining Limits and Using Limit Notation

4. Topic 1.3 · Say it in contextOn a graph, as $x\to 3$ from the left $f(x)$ approaches 5, from the right $f(x)$ approaches 2, and $f(3)=5$. What is ${\lim }_{x\to 3}f(x)$?

   It does not exist. The one-sided limits (5 and 2) differ, and the value $f(3)=5$ does not change that.

   Review: 1.3 Estimating Limit Values from Graphs

5. Topic 1.4 · Say it in contextA table shows $f(1.9)=3.9$, $f(1.99)=3.99$, $f(1.999)=3.999$ and $f(2.1)=4.1$, $f(2.01)=4.01$, $f(2.001)=4.001$. Estimate ${\lim }_{x\to 2}f(x)$. What does the table not tell you?

   About 4, since the values from both sides approach 4. A table only estimates: it cannot prove the limit or show behavior between the sampled inputs.

   Review: 1.4 Estimating Limit Values from Tables

6. Topic 1.5 · Work it outGiven ${\lim }_{x\to 3}f(x)=4$ and ${\lim }_{x\to 3}g(x)=-2$, find ${\lim }_{x\to 3}\frac{f(x)+2g(x)}{g(x{)}^2}$.

   Limit of the numerator: $4+2(-2)=0$. Limit of the denominator: $(-2{)}^2=4$, which is not 0, so the quotient rule for limits applies and the answer is $0/4=0$.

   Review: 1.5 Determining Limits Using Algebraic Properties of Limits

7. Topic 1.6 · Work it outFind ${\lim }_{x\to 3}\frac{x^2-9}{x-3}$. Direct substitution gives $0/0$.

   Factor and cancel: $\frac{(x-3)(x+3)}{x-3}=x+3$ for $x\ne 3$. The limit is 6.

   Review: 1.6 Determining Limits Using Algebraic Manipulation

8. Topic 1.6 · Work it outFind ${\lim }_{x\to 0}\frac{\sqrt{x+4}-2}{x}$.

   Multiply by the conjugate: $\frac{(x+4)-4}{x(\sqrt{x+4}+2)}=\frac{1}{\sqrt{x+4}+2}$. As $x\to 0$ this approaches $\frac{1}{4}$.

   Review: 1.6 Determining Limits Using Algebraic Manipulation

9. Topic 1.7 · Which one fitsName the first move for each limit: (a) ${\lim }_{x\to 4}\frac{x^2-16}{x-4}$, (b) ${\lim }_{x\to 0}\frac{\sqrt{x+9}-3}{x}$, (c) ${\lim }_{x\to 5}\frac{x^2+1}{x-1}$.

   (a) Factor and cancel: 8. (b) Multiply by the conjugate: $\frac{1}{6}$. (c) Direct substitution, since the denominator is not 0: $\frac{26}{4}=\frac{13}{2}$.

   Review: 1.7 Selecting Procedures for Determining Limits

10. Topic 1.8 · Check the conditionsWhat must be true about $g$, $h$ and $f$ near $x=a$ to use the Squeeze Theorem and conclude ${\lim }_{x\to a}h(x)=L$?

    $g(x)\le h(x)\le f(x)$ near $a$ (except possibly at $a$), and ${\lim }_{x\to a}g(x)={\lim }_{x\to a}f(x)=L$. Both outer limits must exist and be equal.

    Review: 1.8 Determining Limits Using the Squeeze Theorem

11. Topic 1.8 · Work it outUse $-|x|\le x\sin \frac{1}{x}\le |x|$ to find ${\lim }_{x\to 0}x\sin \frac{1}{x}$.

    Both $-|x|$ and $|x|$ approach 0 as $x\to 0$, so by the Squeeze Theorem the limit is 0.

    Review: 1.8 Determining Limits Using the Squeeze Theorem

12. Topic 1.9 · Say it in contextThe same limit is given as a graph, a table and a formula. Which representation can give the exact value, and which usually only estimates it?

    An algebraic formula can give the exact value. A graph or a table usually only estimates, because it shows a finite set of points and cannot rule out a change between them.

    Review: 1.9 Connecting Multiple Representations of Limits

13. Topic 1.10 · Which one fitsName the discontinuity at the given $x$: (a) $\frac{x^2-1}{x-1}$ at $x=1$; (b) a function with left limit 2 and right limit 5 at $x=3$; (c) $\frac{1}{x-2}$ at $x=2$.

    (a) Removable: the limit is 2 but $f(1)$ is not defined. (b) Jump: the one-sided limits differ. (c) Infinite: a vertical asymptote.

    Review: 1.10 Exploring Types of Discontinuities

14. Topic 1.11 · Check the conditionsWhat three things must hold for $f$ to be continuous at $x=c$?

    $f(c)$ is defined, ${\lim }_{x\to c}f(x)$ exists, and the limit equals $f(c)$. If any one fails, $f$ is discontinuous at $c$.

    Review: 1.11 Defining Continuity at a Point

15. Topic 1.12 · Work it outLet $f(x)=x^2-1$ for $x<2$ and $f(x)=kx+1$ for $x\ge 2$. Find $k$ so $f$ is continuous at $x=2$.

    Left limit: $4-1=3$. Right side: $f(2)=2k+1$. Set $2k+1=3$, so $k=1$.

    Review: 1.12 Confirming Continuity over an Interval

16. Topic 1.12 · Check the conditionsWhat does it take for $f$ to be continuous on a closed interval $[a,b]$?

    Continuous at every point of $(a,b)$, continuous from the right at $a$, and continuous from the left at $b$.

    Review: 1.12 Confirming Continuity over an Interval

17. Topic 1.13 · Work it out$f(x)=\frac{x^2-x-6}{x-3}$ for $x\ne 3$. What value of $f(3)$ removes the discontinuity?

    For $x\ne 3$, $f(x)=\frac{(x-3)(x+2)}{x-3}=x+2$, so the limit at 3 is 5. Define $f(3)=5$.

    Review: 1.13 Removing Discontinuities

18. Topic 1.14 · Say it in context${\lim }_{x\to 2^{-}}f(x)=-\infty$ and ${\lim }_{x\to 2^{+}}f(x)=+\infty$. What does this say about the graph, and does the limit exist?

    The graph has a vertical asymptote at $x=2$. The two-sided limit does not exist: $\pm \infty$ describes unbounded behavior, not a real number.

    Review: 1.14 Connecting Infinite Limits and Vertical Asymptotes

19. Topic 1.14 · Spot the mistakeA student says $\frac{x+1}{x^2-x-2}$ has vertical asymptotes at $x=-1$ and $x=2$. Correct it.

    The denominator is $(x-2)(x+1)$, and the factor $x+1$ cancels. So there is a hole at $x=-1$ and only one vertical asymptote, at $x=2$.

    Review: 1.14 Connecting Infinite Limits and Vertical Asymptotes

20. Topic 1.15 · Which one fitsFor a rational function, what is the horizontal asymptote when the numerator degree is less than, equal to, or greater than the denominator degree?

    Less: $y=0$. Equal: $y$ equals the ratio of the leading coefficients. Greater: no horizontal asymptote, since the limit at infinity is infinite.

    Review: 1.15 Connecting Limits at Infinity and Horizontal Asymptotes

21. Topic 1.15 · Work it outFind ${\lim }_{x\to \infty }\frac{3x^2-x}{5x^2+2}$ and name the horizontal asymptote.

    Divide by $x^2$: $\frac{3-1/x}{5+2/x^2}\to \frac{3}{5}$. The horizontal asymptote is $y=\frac{3}{5}$.

    Review: 1.15 Connecting Limits at Infinity and Horizontal Asymptotes

22. Topic 1.16 · Check the conditionsWhat must be true to use the Intermediate Value Theorem to conclude $f(c)=k$ for some $c$ in $(a,b)$?

    $f$ is continuous on $[a,b]$ and $k$ lies strictly between $f(a)$ and $f(b)$. Without continuity there is no guarantee.

    Review: 1.16 Working with the Intermediate Value Theorem (IVT)

23. Topic 1.16 · Work it out$f$ is continuous on $[\mathrm{1,4}]$ with $f(1)=-2$ and $f(4)=5$. Does the IVT guarantee a solution of $f(x)=0$ in $(\mathrm{1,4})$? Of $f(x)=7$?

    Yes for 0, since $-2<0<5$. No for 7, since 7 is not between $-2$ and 5, so the theorem says nothing.

    Review: 1.16 Working with the Intermediate Value Theorem (IVT)

Unit 2: Differentiation: Definition and Fundamental Properties

1. Topic 2.1 · Say it in context$s(t)$ is a car's position in meters after $t$ seconds, and $\frac{s(5)-s(2)}{5-2}=4$. Interpret it with units.

   The average velocity of the car on the interval from $t=2$ to $t=5$ seconds is 4 meters per second.

   Review: 2.1 Defining Average and Instantaneous Rates of Change at a Point

2. Topic 2.1 · Work it outUse a limit of a difference quotient to find the instantaneous rate of change of $f(x)=x^2$ at $x=3$.

   ${\lim }_{h\to 0}\frac{(3+h{)}^2-9}{h}={\lim }_{h\to 0}(6+h)=6$.

   Review: 2.1 Defining Average and Instantaneous Rates of Change at a Point

3. Topic 2.2 · Know the termWrite the limit definition of $f^{\prime }(x)$.

   $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$. At a point $a$ it can be written $f^{\prime }(a)={\lim }_{x\to a}\frac{f(x)-f(a)}{x-a}$.

   Review: 2.2 Defining the Derivative of a Function and Using Derivative Notation

4. Topic 2.2 · Work it outUse the limit definition to find $f^{\prime }(x)$ for $f(x)=x^2-x$.

   $\frac{(x+h{)}^2-(x+h)-(x^2-x)}{h}=\frac{2xh+h^2-h}{h}=2x+h-1$. As $h\to 0$, $f^{\prime }(x)=2x-1$.

   Review: 2.2 Defining the Derivative of a Function and Using Derivative Notation

5. Topic 2.3 · Work it out$f(1.9)=6.4$, $f(2.0)=7.0$, $f(2.1)=7.7$. Estimate $f^{\prime }(2)$ using the best interval.

   Use the symmetric interval: $\frac{7.7-6.4}{2.1-1.9}=6.5$. The one-sided estimates, 6 and 7, bracket it.

   Review: 2.3 Estimating Derivatives of a Function at a Point

6. Topic 2.4 · Check the conditionsLet $f(x)=|x-2|$. Is $f$ continuous at 2? Differentiable at 2?

   Continuous: yes. Differentiable: no. The graph has a corner at 2, with slope $-1$ on the left and $1$ on the right, so the derivative does not exist there.

   Review: 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

7. Topic 2.4 · Spot the mistakeA student says: "$f$ is continuous at $c$, so $f^{\prime }(c)$ exists." Give a counterexample and name other ways a derivative can fail.

   $f(x)=|x|$ is continuous at 0 but has a corner. A derivative also fails at a cusp, a vertical tangent, or a discontinuity. Differentiable implies continuous, not the reverse.

   Review: 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

8. Topic 2.5 · Work it outDifferentiate $f(x)={\displaystyle \frac{4}{\sqrt{x}}}$.

   Rewrite as $4x^{-1/2}$. Then $f^{\prime }(x)=4\cdot (-\frac{1}{2})x^{-3/2}=-2x^{-3/2}$.

   Review: 2.5 Applying the Power Rule

9. Topic 2.6 · Work it outFind $f^{\prime }(1)$ for $f(x)=3x^4-2x^2+7x-9$.

   $f^{\prime }(x)=12x^3-4x+7$, so $f^{\prime }(1)=12-4+7=15$.

   Review: 2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple

10. Topic 2.7 · Know the termGive the derivatives of $\sin x$, $\cos x$, $e^x$ and $\ln x$.

    $\cos x$, $-\sin x$, $e^x$, and $\frac{1}{x}$ (for $x>0$).

    Review: 2.7 Derivatives of cos x, sin x, eˣ, and ln x

11. Topic 2.7 · Work it outDifferentiate $f(x)=3e^x-2\sin x+\ln x$.

    $f^{\prime }(x)=3e^x-2\cos x+\frac{1}{x}$.

    Review: 2.7 Derivatives of cos x, sin x, eˣ, and ln x

12. Topic 2.8 · Work it outDifferentiate $f(x)=x^2{e}^x$.

    Product rule: $f^{\prime }(x)=2x{e}^x+x^2{e}^x=x{e}^x(x+2)$.

    Review: 2.8 The Product Rule

13. Topic 2.8 · Spot the mistakeA student says the derivative of $x^3\sin x$ is $3x^2\cos x$. What went wrong?

    The student multiplied the derivatives. The product rule gives $3x^2\sin x+x^3\cos x$.

    Review: 2.8 The Product Rule

14. Topic 2.9 · Work it outDifferentiate $f(x)={\displaystyle \frac{x^2}{x+1}}$.

    Quotient rule: $\frac{2x(x+1)-x^2}{(x+1{)}^2}=\frac{x^2+2x}{(x+1{)}^2}$.

    Review: 2.9 The Quotient Rule

15. Topic 2.9 · Spot the mistakeA student writes ${\left(\frac{f}{g}\right)}^{\prime }=\frac{f{g}^{\prime }-f^{\prime }g}{g^2}$. Correct it.

    The numerator is $f^{\prime }g-f{g}^{\prime }$: low times derivative of high, minus high times derivative of low. The order matters because the two terms subtract.

    Review: 2.9 The Quotient Rule

16. Topic 2.10 · Know the termGive the derivatives of $\tan x$, $\cot x$, $\sec x$ and $\csc x$.

    ${\sec }^{2}x$, $-{\csc }^{2}x$, $\sec x\tan x$, and $-\csc x\cot x$.

    Review: 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

17. Topic 2.10 · Work it outDifferentiate $f(x)=x\tan x$.

    Product rule: $f^{\prime }(x)=\tan x+x{\sec }^{2}x$.

    Review: 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

1. Topic 3.1 · Work it outDifferentiate $f(x)=(3x^2+1{)}^5$.

   Chain rule: $5(3x^2+1{)}^4\cdot 6x=30x(3x^2+1{)}^4$.

   Review: 3.1 The Chain Rule

2. Topic 3.1 · Work it outDifferentiate $f(x)=\sin \left(e^{2x}\right)$.

   Work from the outside in: $\cos (e^{2x})\cdot e^{2x}\cdot 2=2e^{2x}\cos (e^{2x})$.

   Review: 3.1 The Chain Rule

3. Topic 3.1 · Spot the mistakeA student says $\frac{d}{dx}\ln (x^2+1)=\frac{1}{x^2+1}$. What is missing?

   The chain rule factor, the derivative of the inside: $\frac{2x}{x^2+1}$.

   Review: 3.1 The Chain Rule

4. Topic 3.1 · Work it out$h(x)=f(g(x))$, with $g(2)=3$, $g^{\prime }(2)=-4$ and $f^{\prime }(3)=5$. Find $h^{\prime }(2)$.

   $h^{\prime }(2)=f^{\prime }(g(2))\cdot g^{\prime }(2)=f^{\prime }(3)\cdot (-4)=-20$.

   Review: 3.1 The Chain Rule

5. Topic 3.2 · Work it outFor $x^2+y^2=25$, find $\frac{dy}{dx}$ and the slope of the tangent at $(\mathrm{3,4})$.

   $2x+2y\frac{dy}{dx}=0$, so $\frac{dy}{dx}=-\frac{x}{y}$. At $(\mathrm{3,4})$ the slope is $-\frac{3}{4}$.

   Review: 3.2 Implicit Differentiation

6. Topic 3.2 · Work it outFind $\frac{dy}{dx}$ for $x^{2}y+y^3=5$.

   Product rule on $x^{2}y$: $2xy+x^2{y}^{\prime }+3y^2{y}^{\prime }=0$. Solve: $y^{\prime }=-\frac{2xy}{x^2+3y^2}$.

   Review: 3.2 Implicit Differentiation

7. Topic 3.3 · Work it out$f$ is invertible with $f(2)=5$ and $f^{\prime }(2)=4$. Find ${\left(f^{-1}\right)}^{\prime }(5)$.

   ${\left(f^{-1}\right)}^{\prime }(5)=\frac{1}{f^{\prime }(f^{-1}(5))}=\frac{1}{f^{\prime }(2)}=\frac{1}{4}$.

   Review: 3.3 Differentiating Inverse Functions

8. Topic 3.4 · Know the termGive the derivatives of $\arcsin x$, $\arccos x$ and $\arctan x$.

   $\frac{1}{\sqrt{1-x^2}}$, $-\frac{1}{\sqrt{1-x^2}}$, and $\frac{1}{1+x^2}$.

   Review: 3.4 Differentiating Inverse Trigonometric Functions

9. Topic 3.4 · Work it outDifferentiate $f(x)=\arctan (3x)$.

   Chain rule: $\frac{1}{1+(3x{)}^2}\cdot 3=\frac{3}{1+9x^2}$.

   Review: 3.4 Differentiating Inverse Trigonometric Functions

10. Topic 3.5 · Which one fitsName the first rule you use: (a) $x^2\sin x$, (b) $\sin (x^2)$, (c) $\frac{\sin x}{x^2}$, (d) $(\sin x{)}^2$.

    (a) Product rule. (b) Chain rule, inside function $x^2$. (c) Quotient rule. (d) Chain rule, power of the inside function $\sin x$.

    Review: 3.5 Selecting Procedures for Calculating Derivatives

11. Topic 3.6 · Work it outFind $f^{\prime \prime }(x)$ for $f(x)=x^4-3x^2+2$.

    $f^{\prime }(x)=4x^3-6x$, so $f^{\prime \prime }(x)=12x^2-6$.

    Review: 3.6 Calculating Higher-Order Derivatives

12. Topic 3.6 · Work it outFind $\frac{d^{2}y}{d{x}^2}$ for $y=e^{2x}$ and for $y=\ln x$.

    For $e^{2x}$: $2e^{2x}$, then $4e^{2x}$. For $\ln x$: $\frac{1}{x}$, then $-\frac{1}{x^2}$.

    Review: 3.6 Calculating Higher-Order Derivatives

Unit 4: Contextual Applications of Differentiation

1. Topic 4.1 · Say it in context$C(q)$ is the cost in dollars of making $q$ units, and $C^{\prime }(100)=12.5$. Interpret it with units.

   At a production level of 100 units, cost is increasing at a rate of 12.50 dollars per unit. That approximates the extra cost of making the 101st unit.

   Review: 4.1 Interpreting the Meaning of the Derivative in Context

2. Topic 4.2 · Work it outA particle has position $x(t)=t^3-6t^2+9t$. At $t=4$, find the velocity and acceleration, and say whether the particle is speeding up.

   $v=3t^2-12t+9$ gives $v(4)=9$. $a=6t-12$ gives $a(4)=12$. Velocity and acceleration are both positive, so the particle is speeding up.

   Review: 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

3. Topic 4.2 · Check the conditionsHow do you tell from $v(t)$ and $a(t)$ whether a particle is speeding up or slowing down?

   Speeding up when $v$ and $a$ have the same sign. Slowing down when they have opposite signs. If $a=0$, the speed is not changing at that instant.

   Review: 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

4. Topic 4.3 · Say it in context$V(t)$ is the volume of water in a tank, in liters, $t$ minutes after noon, and $V^{\prime }(5)=-2$. Interpret it.

   Five minutes after noon, the volume of water is decreasing at 2 liters per minute.

   Review: 4.3 Rates of Change in Applied Contexts Other Than Motion

5. Topic 4.4 · Work it outThe volume of a sphere is $V=\frac{4}{3}\pi r^3$, and both $V$ and $r$ change with time. Write $\frac{dV}{dt}$ in terms of $\frac{dr}{dt}$.

   Differentiate with respect to $t$, treating $r$ as a function of $t$: $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$.

   Review: 4.4 Introduction to Related Rates

6. Topic 4.4 · Spot the mistakeA student sets $r=5$ in $V=\frac{4}{3}\pi r^3$ before differentiating and gets $\frac{dV}{dt}=0$. What is wrong?

   The radius changes with time, so substituting a number first treats it as a constant. Differentiate first, then substitute the values at the instant.

   Review: 4.4 Introduction to Related Rates

7. Topic 4.5 · Work it outA 10 ft ladder slides down a wall. Its bottom moves away from the wall at 2 ft/s. How fast is the top moving down when the bottom is 6 ft from the wall?

   $x^2+y^2=100$. At $x=6$, $y=8$. $2x{x}^{\prime }+2y{y}^{\prime }=0$ gives $12(2)+16y^{\prime }=0$, so $y^{\prime }=-1.5$. The top is falling at 1.5 ft/s.

   Review: 4.5 Solving Related Rates Problems

8. Topic 4.5 · Work it outThe area of a circular oil slick grows at 12 cm${}^2$/s. How fast is the radius growing when $r=3$ cm?

   $A=\pi r^2$, so $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$. Then $12=6\pi \frac{dr}{dt}$, so $\frac{dr}{dt}=\frac{2}{\pi }$ cm/s.

   Review: 4.5 Solving Related Rates Problems

9. Topic 4.6 · Work it outUse the linearization of $f(x)=\sqrt{x}$ at $x=4$ to estimate $\sqrt{4.1}$.

   $L(x)=2+\frac{1}{4}(x-4)$. Then $L(4.1)=2+0.025=2.025$.

   Review: 4.6 Approximating Values of a Function Using Local Linearity and Linearization

10. Topic 4.6 · Say it in context$f$ is concave up near $x=a$. Does the tangent line approximation at $a$ over- or underestimate $f$ nearby?

    It underestimates. A concave up graph lies above its tangent lines.

    Review: 4.6 Approximating Values of a Function Using Local Linearity and Linearization

11. Topic 4.7 · Check the conditionsWhen may you use L'Hospital's Rule on ${\lim }_{x\to a}\frac{f(x)}{g(x)}$, and what do you then compute?

    Only when substitution gives $\frac{0}{0}$ or $\frac{\pm \infty }{\pm \infty }$. Then the limit equals $\lim \frac{f^{\prime }(x)}{g^{\prime }(x)}$ if that exists. Differentiate top and bottom separately, not as a quotient.

    Review: 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

12. Topic 4.7 · Work it outFind ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}$.

    The form is $\frac{0}{0}$. First pass: $\frac{\sin x}{2x}$, still $\frac{0}{0}$. Second pass: $\frac{\cos x}{2}\to \frac{1}{2}$.

    Review: 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

13. Topic 4.7 · Spot the mistakeA student uses L'Hospital's Rule on ${\lim }_{x\to 0}\frac{x+2}{x+1}$ and gets $\frac{1}{1}=1$. What is wrong?

    The form is $\frac{2}{1}$, not indeterminate, so the rule does not apply. Direct substitution gives 2.

    Review: 4.7 Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

Unit 5: Analytical Applications of Differentiation

1. Topic 5.1 · Check the conditionsState the Mean Value Theorem, including its conditions.

   If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c$ in $(a,b)$ has $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$.

   Review: 5.1 Using the Mean Value Theorem

2. Topic 5.1 · Work it outFind the value of $c$ the Mean Value Theorem guarantees for $f(x)=x^3$ on $[\mathrm{0,2}]$.

   The average rate is $\frac{8-0}{2}=4$. Solve $3c^2=4$: $c=\frac{2}{\sqrt{3}}\approx 1.155$, which is in $(\mathrm{0,2})$.

   Review: 5.1 Using the Mean Value Theorem

3. Topic 5.1 · Say it in contextA car travels 120 miles in 2 hours. What does the Mean Value Theorem guarantee?

   At some moment during the trip the car's instantaneous speed was exactly 60 mph, assuming its position is differentiable.

   Review: 5.1 Using the Mean Value Theorem

4. Topic 5.2 · Know the termWhat is a critical point of $f$?

   A point $c$ in the domain of $f$ where $f^{\prime }(c)=0$ or $f^{\prime }(c)$ does not exist.

   Review: 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

5. Topic 5.2 · Check the conditionsWhat does the Extreme Value Theorem guarantee, and what must be true?

   If $f$ is continuous on a closed interval $[a,b]$, then $f$ has both an absolute maximum and an absolute minimum on it. Drop either condition and the guarantee is gone.

   Review: 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

6. Topic 5.2 · Spot the mistakeA student says every critical point is a relative maximum or minimum. Give a counterexample.

   For $f(x)=x^3$, $f^{\prime }(0)=0$, so 0 is a critical point, but $f$ is increasing on both sides of 0. It is neither.

   Review: 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

7. Topic 5.3 · Work it outOn what intervals is $f(x)=x^3-3x^2$ increasing?

   $f^{\prime }(x)=3x(x-2)$, positive when $x<0$ or $x>2$. So $f$ is increasing on $(-\infty ,0)$ and $(2,\infty )$.

   Review: 5.3 Determining Intervals on Which a Function Is Increasing or Decreasing

8. Topic 5.4 · Work it out$f^{\prime }(x)=(x-1{)}^2(x+2)$. Classify each critical point.

   At $x=-2$, $f^{\prime }$ changes from negative to positive: relative minimum. At $x=1$, the factor $(x-1{)}^2$ never changes sign, so $f^{\prime }$ does not change sign: neither.

   Review: 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema

9. Topic 5.5 · Work it outFind the absolute extrema of $f(x)=x^3-3x$ on $[-\mathrm{2,3}]$.

   Candidates: endpoints $-2$, $3$ and critical points $\pm 1$. Values: $-2$, $2$, $-2$, $18$ at $x=-2,-\mathrm{1,1,3}$. Absolute max 18 at $x=3$. Absolute min $-2$ at $x=-2$ and $x=1$.

   Review: 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

10. Topic 5.6 · Check the conditionsHow does $f^{\prime \prime }$ tell you about concavity, and what makes a point an inflection point?

    $f^{\prime \prime }>0$: concave up. $f^{\prime \prime }<0$: concave down. An inflection point is where concavity changes, so $f^{\prime \prime }$ must change sign. $f^{\prime \prime }=0$ alone is not enough.

    Review: 5.6 Determining Concavity of Functions over Their Domains

11. Topic 5.6 · Work it outFind the inflection points of $f(x)=x^4-6x^2$.

    $f^{\prime \prime }(x)=12x^2-12=12(x-1)(x+1)$, which changes sign at both $x=-1$ and $x=1$. Both are inflection points.

    Review: 5.6 Determining Concavity of Functions over Their Domains

12. Topic 5.6 · Spot the mistakeA student says $f(x)=x^4$ has an inflection point at 0 because $f^{\prime \prime }(0)=0$. Is that right?

    No. $f^{\prime \prime }(x)=12x^2\ge 0$ never changes sign, so $f$ is concave up on both sides of 0. There is no inflection point.

    Review: 5.6 Determining Concavity of Functions over Their Domains

13. Topic 5.7 · Check the conditionsState the Second Derivative Test for a critical point $c$ where $f^{\prime }(c)=0$.

    If $f^{\prime \prime }(c)>0$, relative minimum. If $f^{\prime \prime }(c)<0$, relative maximum. If $f^{\prime \prime }(c)=0$, the test is inconclusive, so use the First Derivative Test.

    Review: 5.7 Using the Second Derivative Test to Determine Extrema

14. Topic 5.7 · Work it outClassify the critical points of $f(x)=x^3-12x$ with the Second Derivative Test.

    $f^{\prime }(x)=3x^2-12$, so $x=\pm 2$. $f^{\prime \prime }(x)=6x$. $f^{\prime \prime }(-2)=-12<0$: relative maximum. $f^{\prime \prime }(2)=12>0$: relative minimum.

    Review: 5.7 Using the Second Derivative Test to Determine Extrema

15. Topic 5.8 · Say it in context$f$ is increasing before $x=2$, has a horizontal tangent at $x=2$, and is decreasing after. What does the graph of $f^{\prime }$ do at $x=2$?

    $f^{\prime }$ crosses the $x$-axis at $x=2$, going from positive to negative.

    Review: 5.8 Sketching Graphs of Functions and Their Derivatives

16. Topic 5.9 · Which one fitsMatch each feature of $f$ with the behavior of $f^{\prime }$ and $f^{\prime \prime }$: increasing, concave up, an inflection point.

    Increasing: $f^{\prime }>0$. Concave up: $f^{\prime \prime }>0$, so $f^{\prime }$ is increasing. Inflection point of $f$: $f^{\prime \prime }$ changes sign, so $f^{\prime }$ has a local extremum.

    Review: 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative

17. Topic 5.9 · Say it in contextOn $(\mathrm{0,3})$ the graph of $f^{\prime }$ is above the $x$-axis and falling. What can you say about $f$ there?

    $f^{\prime }>0$ means $f$ is increasing. $f^{\prime }$ falling means $f^{\prime \prime }<0$, so $f$ is concave down. So $f$ rises, bending downward.

    Review: 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative

18. Topic 5.10 · Work it outA rectangle has perimeter 40. Write its area as a function of one variable, and say what values of $x$ make sense.

    With sides $x$ and $y$: $2x+2y=40$ gives $y=20-x$. So $A(x)=x(20-x)$ on $0<x<20$. The endpoints give area 0, so $[\mathrm{0,20}]$ is also acceptable.

    Review: 5.10 Introduction to Optimization Problems

19. Topic 5.11 · Work it outSquares of side $x$ are cut from the corners of a 12 by 12 sheet and the sides are folded up into an open box. Find the greatest volume.

    $V=x(12-2x{)}^2$ on $0<x<6$. $V^{\prime }=(12-2x)(12-6x)=0$ gives $x=2$. $V(2)=128$, and $V$ is 0 at both ends, so the maximum is 128.

    Review: 5.11 Solving Optimization Problems

20. Topic 5.11 · Spot the mistakeAfter solving $V^{\prime }=0$, a student reports the critical number as the answer. What is missing?

    A justification that it is a maximum, not a minimum or neither: use the First or Second Derivative Test, or compare with the endpoints (Candidates Test). Then answer what was asked, the greatest value of the quantity, not just the critical number.

    Review: 5.11 Solving Optimization Problems

21. Topic 5.12 · Work it outFor $x^2+y^2=25$, where are the tangent lines horizontal? Vertical?

    $\frac{dy}{dx}=-\frac{x}{y}$. Horizontal where $x=0$: points $(0,\pm 5)$. Vertical where $y=0$: points $(\pm \mathrm{5,0})$.

    Review: 5.12 Exploring Behaviors of Implicit Relations

Unit 6: Integration and Accumulation of Change

1. Topic 6.1 · Say it in contextWater flows into a tank at rate $r(t)$ gallons per minute. What does ${\int }_0^{10}r(t)dt$ represent?

   The total number of gallons that flowed in during the first 10 minutes: the accumulated change.

   Review: 6.1 Exploring Accumulations of Change

2. Topic 6.2 · Work it outFind the left and right Riemann sums for $f(x)=x^2$ on $[\mathrm{0,4}]$ with 4 equal subintervals.

   $\mathrm{\Delta }x=1$. Left: $0+1+4+9=14$. Right: $1+4+9+16=30$. For an increasing function the left sum underestimates and the right sum overestimates.

   Review: 6.2 Approximating Areas with Riemann Sums

3. Topic 6.2 · Work it out$r(t)$ is measured at $t=\mathrm{0,2,5,6}$ with values $\mathrm{4,6,3,5}$. Find the right Riemann sum and the trapezoidal sum for ${\int }_0^{6}r(t)dt$.

   Widths 2, 3, 1. Right: $6(2)+3(3)+5(1)=26$. Trapezoids: $\frac{4+6}{2}(2)+\frac{6+3}{2}(3)+\frac{3+5}{2}(1)=10+13.5+4=27.5$.

   Review: 6.2 Approximating Areas with Riemann Sums

4. Topic 6.3 · Know the termWrite the definite integral as the limit of a Riemann sum.

   ${\int }_a^{b}f(x)dx={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(x_i)\mathrm{\Delta }x$, where $\mathrm{\Delta }x=\frac{b-a}{n}$ and $x_i$ is a point in the $i$th subinterval, such as the right endpoint $a+i\mathrm{\Delta }x$.

   Review: 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

5. Topic 6.3 · Work it outWrite ${\lim }_{n\to \infty }{\sum }_{i=1}^n{(2+\frac{3i}{n})}^2\cdot \frac{3}{n}$ as a definite integral.

   $\mathrm{\Delta }x=\frac{3}{n}$ and $x_i=2+\frac{3i}{n}$, so the interval is $[\mathrm{2,5}]$ and $f(x)=x^2$. The integral is ${\int }_2^5{x}^{2}dx$.

   Review: 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

6. Topic 6.4 · Work it outLet $F(x)={\int }_1^x(t^2+1)dt$. Find $F^{\prime }(x)$ and $F(1)$.

   By the Fundamental Theorem, $F^{\prime }(x)=x^2+1$. And $F(1)=0$, since the interval has zero width.

   Review: 6.4 The Fundamental Theorem of Calculus and Accumulation Functions

7. Topic 6.4 · Work it outLet $G(x)={\int }_0^{x^2}\sin tdt$. Find $G^{\prime }(x)$.

   Chain rule with the Fundamental Theorem: $G^{\prime }(x)=\sin (x^2)\cdot 2x$.

   Review: 6.4 The Fundamental Theorem of Calculus and Accumulation Functions

8. Topic 6.5 · Say it in context$F(x)={\int }_0^{x}f(t)dt$. $f$ is positive on $(\mathrm{0,3})$, negative on $(\mathrm{3,6})$. What does $F$ do at $x=3$?

   $F^{\prime }=f$ changes from positive to negative at 3, so $F$ increases and then decreases: a relative maximum at $x=3$.

   Review: 6.5 Interpreting the Behavior of Accumulation Functions Involving Area

9. Topic 6.6 · Work it out${\int }_0^{5}f=7$ and ${\int }_0^{2}f=3$. Find ${\int }_2^{5}f(x)dx$ and ${\int }_5^{0}f(x)dx$.

   ${\int }_2^{5}f=7-3=4$. Reversing the limits flips the sign: ${\int }_5^{0}f=-7$.

   Review: 6.6 Applying Properties of Definite Integrals

10. Topic 6.6 · Work it outUse geometry to evaluate ${\int }_{-3}^3\sqrt{9-x^2}dx$.

    The graph is the upper half of a circle of radius 3. The area is $\frac{1}{2}\pi (3{)}^2=\frac{9\pi }{2}$.

    Review: 6.6 Applying Properties of Definite Integrals

11. Topic 6.7 · Work it outEvaluate ${\int }_0^2(3x^2-2x)dx$.

    An antiderivative is $x^3-x^2$. Then $(8-4)-(0)=4$.

    Review: 6.7 The Fundamental Theorem of Calculus and Definite Integrals

12. Topic 6.7 · Say it in context$V(t)$ is the volume of water in a tank, $V^{\prime }(t)=r(t)$, and $V(0)=10$. Write an expression for $V(5)$.

    Net change: $V(5)=V(0)+{\int }_0^{5}r(t)dt=10+{\int }_0^{5}r(t)dt$.

    Review: 6.7 The Fundamental Theorem of Calculus and Definite Integrals

13. Topic 6.7 · Spot the mistakeA student computes ${\int }_{-1}^1\frac{1}{x^2}dx={[-\frac{1}{x}]}_{-1}^1=-2$. Why must this be wrong?

    The integrand is positive, so no integral of it can be negative. The Fundamental Theorem needs $f$ continuous on $[-\mathrm{1,1}]$, and $\frac{1}{x^2}$ is not defined at 0.

    Review: 6.7 The Fundamental Theorem of Calculus and Definite Integrals

14. Topic 6.8 · Work it outFind $\int (4x^3-\frac{2}{x}+e^x)dx$.

    $x^4-2\ln |x|+e^x+C$.

    Review: 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

15. Topic 6.8 · Work it out$f^{\prime }(x)=6x^2-1$ and $f(1)=4$. Find $f(x)$.

    $f(x)=2x^3-x+C$. Then $f(1)=1+C=4$, so $C=3$ and $f(x)=2x^3-x+3$.

    Review: 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

16. Topic 6.9 · Work it outFind $\int 2x\cos \left(x^2\right)dx$.

    Let $u=x^2$, $du=2xdx$. The integral becomes $\int \cos udu=\sin \left(x^2\right)+C$.

    Review: 6.9 Integrating Using Substitution

17. Topic 6.9 · Spot the mistakeTo evaluate ${\int }_0^{2}x(x^2+1{)}^{3}dx$, a student uses $u=x^2+1$ and keeps the limits 0 and 2. What is wrong, and what is the value?

    The limits must change with the variable: $u(0)=1$, $u(2)=5$. Then $\frac{1}{2}{\int }_1^5{u}^{3}du=\frac{1}{2}\cdot \frac{625-1}{4}=78$.

    Review: 6.9 Integrating Using Substitution

18. Topic 6.10 · Work it outFind $\int \frac{x^2}{x+1}dx$.

    Divide: $\frac{x^2}{x+1}=x-1+\frac{1}{x+1}$. The integral is $\frac{x^2}{2}-x+\ln |x+1|+C$.

    Review: 6.10 Integrating Functions Using Long Division and Completing the Square

19. Topic 6.10 · Work it outFind $\int \frac{1}{x^2+4x+13}dx$.

    Complete the square: $x^2+4x+13=(x+2{)}^2+9$. The integral is $\frac{1}{3}\arctan \frac{x+2}{3}+C$.

    Review: 6.10 Integrating Functions Using Long Division and Completing the Square

20. Topic 6.14 · Which one fitsChoose a technique: (a) $\int \frac{x}{x^2+1}dx$, (b) $\int \frac{x^2+1}{x}dx$, (c) $\int \frac{1}{x^2+9}dx$, (d) $\int \frac{x^2}{x-1}dx$.

    (a) Substitution, $u=x^2+1$. (b) Split the fraction into $x+\frac{1}{x}$. (c) Inverse tangent form, $\frac{1}{3}\arctan \frac{x}{3}$. (d) Long division.

    Review: 6.14 Selecting Techniques for Antidifferentiation

Unit 7: Differential Equations

1. Topic 7.1 · Say it in contextWrite a differential equation for each: (a) a tank's volume $V$ changes at a rate equal to 3 minus twice $V$; (b) the rate of change of $P$ is proportional to $P$.

   (a) $\frac{dV}{dt}=3-2V$. (b) $\frac{dP}{dt}=kP$, for some constant $k$.

   Review: 7.1 Modeling Situations with Differential Equations

2. Topic 7.2 · Work it outIs $y=e^{2x}$ a solution of $y^{\prime \prime }-y^{\prime }-2y=0$?

   $y^{\prime }=2e^{2x}$ and $y^{\prime \prime }=4e^{2x}$. Then $4e^{2x}-2e^{2x}-2e^{2x}=0$. Yes.

   Review: 7.2 Verifying Solutions for Differential Equations

3. Topic 7.3 · Work it outFor $\frac{dy}{dx}=x-y$, find the slope of the slope field segments at $(\mathrm{2,1})$ and at $(\mathrm{0,3})$.

   At $(\mathrm{2,1})$: $2-1=1$. At $(\mathrm{0,3})$: $0-3=-3$.

   Review: 7.3 Sketching Slope Fields

4. Topic 7.3 · Say it in contextIn a slope field for $\frac{dy}{dx}$ that depends only on $y$, how do the segments along any one horizontal line compare?

   They all have the same slope, so they are parallel. If the equation depends only on $x$, the segments along each vertical line match instead.

   Review: 7.3 Sketching Slope Fields

5. Topic 7.4 · Say it in context$\frac{dy}{dx}=y(3-y)$ with $y(0)=1$. Without solving, what is ${\lim }_{x\to \infty }y$?

   For $0<y<3$, $\frac{dy}{dx}>0$, so $y$ increases. It cannot cross the equilibrium $y=3$, so $y$ approaches 3. The limit is 3.

   Review: 7.4 Reasoning Using Slope Fields

6. Topic 7.6 · Work it outFind the general solution of $\frac{dy}{dx}=\frac{x}{y}$.

   Separate: $ydy=xdx$. Integrate: $\frac{y^2}{2}=\frac{x^2}{2}+C$, so $y^2=x^2+C$.

   Review: 7.6 Finding General Solutions Using Separation of Variables

7. Topic 7.6 · Spot the mistakeA student integrates $\frac{dy}{dx}=2xy$ and writes $y=x^{2}y+C$. What is wrong, and what is the right solution?

   You cannot integrate a term containing $y$ with respect to $x$. Separate: $\frac{dy}{y}=2xdx$, so $\ln |y|=x^2+C$ and $y=A{e}^{x^2}$.

   Review: 7.6 Finding General Solutions Using Separation of Variables

8. Topic 7.7 · Work it outSolve $\frac{dy}{dx}=\frac{x^2}{y^2}$ with $y(0)=2$.

   $y^{2}dy=x^{2}dx$ gives $\frac{y^3}{3}=\frac{x^3}{3}+C$, so $y^3=x^3+3C$. Using $y(0)=2$: $8=3C$. So $y=\sqrt[3]{x^3+8}$.

   Review: 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables

9. Topic 7.8 · Say it in contextIf $\frac{dP}{dt}=kP$, what is $P(t)$, and what does the sign of $k$ tell you?

   $P(t)=P_0{e}^{kt}$, where $P_0=P(0)$. If $k>0$ the quantity grows exponentially. If $k<0$ it decays.

   Review: 7.8 Exponential Models with Differential Equations

10. Topic 7.8 · Work it outA population grows at a rate proportional to its size. It is 100 at $t=0$ and 200 at $t=3$. Find $P(t)$.

    $P=100e^{kt}$. From $200=100e^{3k}$, $k=\frac{\ln 2}{3}$. So $P(t)=100e^{(\ln 2)t/3}=100\cdot 2^{t/3}$.

    Review: 7.8 Exponential Models with Differential Equations

11. Topic 7.8 · Work it outSolve $\frac{dy}{dt}=2(y-5)$ with $y(0)=8$.

    Let $u=y-5$, so $\frac{du}{dt}=2u$ and $u=u_0{e}^{2t}$ with $u_0=3$. Then $y=5+3e^{2t}$.

    Review: 7.8 Exponential Models with Differential Equations

Unit 8: Applications of Integration

1. Topic 8.1 · Work it outFind the average value of $f(x)=x^2$ on $[\mathrm{0,3}]$, and a point where $f$ takes that value.

   $\frac{1}{3}{\int }_0^3{x}^{2}dx=\frac{1}{3}\cdot 9=3$. Solve $c^2=3$: $c=\sqrt{3}\approx 1.73$, which is in $[\mathrm{0,3}]$.

   Review: 8.1 Finding the Average Value of a Function on an Interval

2. Topic 8.2 · Work it outA particle has $v(t)=t^2-4t$ on $[\mathrm{0,5}]$. Find the displacement and the total distance traveled.

   Displacement: ${\int }_0^{5}vdt=\frac{125}{3}-50=-\frac{25}{3}$. The velocity is negative on $(\mathrm{0,4})$, so distance is $\frac{32}{3}+\frac{7}{3}=13$.

   Review: 8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

3. Topic 8.3 · Say it in contextPeople enter a park at rate $E(t)$ and leave at rate $L(t)$, in people per hour. There are 500 people at $t=0$. Write the number there at $t=6$, and say when the number is greatest for $0\le t\le 6$.

   $500+{\int }_0^6(E(t)-L(t))dt$. The count is greatest where $E-L$ changes from positive to negative, checking the endpoints as well.

   Review: 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

4. Topic 8.4 · Work it outFind the area between $y=x$ and $y=x^2$ from $x=0$ to $x=1$.

   Top minus bottom: ${\int }_0^1(x-x^2)dx=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.

   Review: 8.4 Finding the Area Between Curves Expressed as Functions of x

5. Topic 8.5 · Work it outFind the area enclosed by $x=y^2$ and $x=y+2$.

   They meet where $y^2=y+2$, so $y=-1$ and $y=2$. Right minus left: ${\int }_{-1}^2(y+2-y^2)dy=\frac{9}{2}$.

   Review: 8.5 Finding the Area Between Curves Expressed as Functions of y

6. Topic 8.6 · Spot the mistakeA student finds the area between $y=x^3$ and $y=x$ on $[-\mathrm{1,1}]$ as ${\int }_{-1}^1(x-x^3)dx=0$. What went wrong?

   The curves cross at $x=0$, so the top curve switches. Split it: $2{\int }_0^1(x-x^3)dx=2(\frac{1}{2}-\frac{1}{4})=\frac{1}{2}$.

   Review: 8.6 Finding the Area Between Curves That Intersect at More Than Two Points

7. Topic 8.7 · Work it outThe base is the region under $y=\sqrt{x}$ on $[\mathrm{0,4}]$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

   Side $\sqrt{x}$, so area $x$. $V={\int }_0^{4}xdx=8$.

   Review: 8.7 Volumes with Cross Sections: Squares and Rectangles

8. Topic 8.8 · Work it outThe base is the region under $y=x$ on $[\mathrm{0,2}]$. Cross sections perpendicular to the $x$-axis are semicircles with diameter in the base. Find the volume.

   Diameter $x$, radius $\frac{x}{2}$, area $\frac{\pi }{2}\cdot \frac{x^2}{4}=\frac{\pi x^2}{8}$. $V=\frac{\pi }{8}{\int }_0^2{x}^{2}dx=\frac{\pi }{8}\cdot \frac{8}{3}=\frac{\pi }{3}$.

   Review: 8.8 Volumes with Cross Sections: Triangles and Semicircles

9. Topic 8.9 · Work it outThe region under $y=\sqrt{x}$ on $[\mathrm{0,4}]$ is revolved about the $x$-axis. Find the volume.

   Discs of radius $\sqrt{x}$: $V=\pi {\int }_0^4(\sqrt{x}{)}^{2}dx=\pi {\int }_0^{4}xdx=8\pi$.

   Review: 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

10. Topic 8.9 · Spot the mistakeA student writes $V=\pi {\int }_0^4\sqrt{x}dx$ for that solid. What is missing?

    The radius must be squared, because a disc has area $\pi r^2$. The integrand is $(\sqrt{x}{)}^2=x$, not $\sqrt{x}$.

    Review: 8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

11. Topic 8.10 · Work it outThe region between $y=x^2$ and $y=4$ for $0\le x\le 2$ is revolved about the line $y=4$. Find the volume.

    Radius $4-x^2$. $V=\pi {\int }_0^2(4-x^2{)}^{2}dx=\pi (32-\frac{64}{3}+\frac{32}{5})=\frac{256\pi }{15}$.

    Review: 8.10 Volume with Disc Method: Revolving Around Other Axes

12. Topic 8.11 · Work it outThe region between $y=x$ and $y=x^2$ is revolved about the $x$-axis. Find the volume.

    Washers, outer radius $x$, inner radius $x^2$: $V=\pi {\int }_0^1(x^2-x^4)dx=\pi (\frac{1}{3}-\frac{1}{5})=\frac{2\pi }{15}$.

    Review: 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis

13. Topic 8.11 · Spot the mistakeFor washers, a student writes $\pi {\int }_0^1(x-x^2{)}^{2}dx$. What is the error?

    That squares the difference of the radii. Square each radius first, then subtract: $\pi \int (R^2-r^2)dx$.

    Review: 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis

14. Topic 8.12 · Work it outThe region between $y=x$ and $y=x^2$ is revolved about the line $y=-1$. Find the volume.

    Outer radius $x+1$, inner radius $x^2+1$. Then $V=\pi {\int }_0^1[(x+1{)}^2-(x^2+1{)}^2]dx$, which simplifies to $\pi {\int }_0^1(2x-x^2-x^4)dx=\frac{7\pi }{15}$.

    Review: 8.12 Volume with Washer Method: Revolving Around Other Axes