AP Calculus AB Contextual Applications of Differentiation — Worked Answer Explanations

Unit 4 · 12% of the AP exam · 12 questions explained

1. Question 1 · Easy

   A particle moves along a number line so that its position is $s(t)=t^3-6t+2$ (feet, $t$ in seconds). What is the velocity at $t=2$?

   • A
     $6$ ft/sCorrect
   • B
     $2$ ft/s

     Why not B: Evaluated $s(2)$ (position) rather than $v(2)=s^{\prime }(2)$.
   • C
     $12$ ft/s

     Why not C: Computed $3t^2$ at $t=2$ without subtracting the derivative of $-6t$.
   • D
     $-6$ ft/s

     Why not D: Used the coefficient of $t$ in $s(t)$ directly rather than differentiating.
   Explanation

   $v(t)=s^{\prime }(t)=3t^2-6$. At $t=2$: $v(2)=3(4)-6=12-6=6$ ft/s.

   Key takeaway

   Velocity is the first derivative of position; substitute after differentiating.

2. Question 2 · Easy

   The radius of a circle is increasing at a rate of $3$ cm/s. At what rate is the area increasing when the radius is $5$ cm?

   • A
     $15\pi$ cm²/s

     Why not A: Used $A=\pi r$ instead of $A=\pi r^2$.
   • B
     $30\pi$ cm²/sCorrect
   • C
     $10\pi$ cm²/s

     Why not C: Forgot to multiply by $\frac{dr}{dt}=3$.
   • D
     $75\pi$ cm²/s

     Why not D: Used $\frac{dA}{dt}=\pi r^2\cdot \frac{dr}{dt}$ instead of $2\pi r\cdot \frac{dr}{dt}$.
   Explanation

   $A=\pi r^2\Rightarrow \frac{dA}{dt}=2\pi r\frac{dr}{dt}$. When $r=5$ and $\frac{dr}{dt}=3$: $\frac{dA}{dt}=2\pi (5)(3)=30\pi$ cm²/s.

   Key takeaway

   Related rates: differentiate the geometric formula with respect to $t$ using the chain rule, then substitute known values.

3. Question 3 · Easy

   Using linear approximation near $x=4$, estimate $\sqrt{4.1}$.

   • A
     $2.025$Correct
   • B
     $2.05$

     Why not B: Used $f^{\prime }(4)=1/2$ without correct denominator $2\sqrt{4}=4$.
   • C
     $2.1$

     Why not C: Applied the increment directly without using the derivative.
   • D
     $4.025$

     Why not D: Added $\Delta x$ to $a$ (the base point) rather than to $f(a)$.
   Explanation

   Linear approximation: $L(x)=f(a)+f^{\prime }(a)(x-a)$. With $f(x)=\sqrt{x}$, $a=4$, $f(4)=2$, $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$, $f^{\prime }(4)=\frac{1}{4}$. So $L(4.1)=2+\frac{1}{4}(0.1)=2.025$.

   Key takeaway

   Linearization uses $L(x)=f(a)+f'(a)(x-a)$; choose $a$ to be a nearby value where $f$ is easy to evaluate.

4. Question 4 · Easy

   The temperature $T$ (°F) of a cooling object satisfies $T^{\prime }(t)=-0.1(T-70)$. At time $t=0$, $T=200$°F. Using the differential equation, what is $T^{\prime }(0)$?

   • A
     $-20$

     Why not A: Used $T=200$ in $-0.1\cdot T$ without subtracting 70.
   • B
     $-13$Correct
   • C
     $-0.1$

     Why not C: Used the constant $-0.1$ as the derivative value.
   • D
     $130$

     Why not D: Computed $T-70=130$ without multiplying by $-0.1$.
   Explanation

   $T^{\prime }(0)=-0.1(T(0)-70)=-0.1(200-70)=-0.1(130)=-13$°F/min.

   Key takeaway

   Evaluate a rate-of-change model by substituting the given initial value directly into the derivative formula.

5. Question 5 · Medium

   The volume of a cube is decreasing at $12$ cm³/s. How fast is the side length decreasing when the side is $2$ cm?

   • A
     $-1$ cm/sCorrect
   • B
     $-2$ cm/s

     Why not B: Divided $12$ by $s=6$ without using the derivative formula $3s^2$.
   • C
     $-3$ cm/s

     Why not C: Divided $12$ by $s^2=4$ without the factor of $3$.
   • D
     $-12$ cm/s

     Why not D: Used $\frac{dV}{dt}$ directly as the side-length rate.
   Explanation

   $V=s^3\Rightarrow \frac{dV}{dt}=3s^2\frac{ds}{dt}$. Given $\frac{dV}{dt}=-12$ and $s=2$: $-12=3(4)\frac{ds}{dt}=12\frac{ds}{dt}$. So $\frac{ds}{dt}=-1$ cm/s.

   Key takeaway

   Related rates for cubes: $\frac{dV}{dt} = 3s^2\frac{ds}{dt}$; solve for the unknown rate after substituting.

6. Question 6 · Medium

   A ladder 10 ft long leans against a vertical wall. The base slides away from the wall at $2$ ft/s. How fast is the top of the ladder sliding down when the base is $6$ ft from the wall?

   • A
     $-\frac{3}{2}$ ft/sCorrect
   • B
     $-2$ ft/s

     Why not B: Used the speed of the base directly without applying related rates.
   • C
     $-\frac{6}{4}$ ft/s

     Why not C: Used $x/y$ without the factor from $dx/dt$.
   • D
     $\frac{3}{2}$ ft/s

     Why not D: Got the correct magnitude but wrong sign; the top moves downward.
   Explanation

   Pythagorean relation: $x^2+y^2=100$. Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. When $x=6$: $y=\sqrt{100-36}=8$. Substituting: $2(6)(2)+2(8)\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{24}{16}=-\frac{3}{2}$ ft/s.

   Key takeaway

   Ladder related-rates problems use the Pythagorean theorem differentiated implicitly with respect to time.

7. Question 7 · Medium

   Apply L'Hôpital's rule to evaluate $\underset{x\to 0}{\lim }\frac{e^x-1-x}{x^2}$.

   • A
     $0$

     Why not A: Applied L'Hôpital only once; the result is still $0/0$.
   • B
     $1$

     Why not B: Applied L'Hôpital once and evaluated too soon.
   • C
     $\frac{1}{2}$Correct
   • D
     Does not exist.

     Why not D: The limit exists and equals $1/2$.
   Explanation

   Check: $0/0$ form. First application: $\frac{e^x-1}{2x}$, still $0/0$ at $x=0$. Second application: $\frac{e^x}{2}\to \frac{1}{2}$ as $x\to 0$.

   Key takeaway

   L'Hôpital's rule may need to be applied more than once if the result remains $0/0$ or $\infty/\infty$.

8. Question 8 · Medium

   The position of a particle is $s(t)=t^3-9t^2+24t$ for $t\ge 0$. At what time(s) is the particle at rest?

   • A
     $t=1$ and $t=8$

     Why not A: Solved $s(t)=0$ (position, not velocity).
   • B
     $t=2$ and $t=4$Correct
   • C
     $t=3$ only

     Why not C: Found the zero of $a(t)=s^{\prime \prime }(t)$, not $v(t)=s^{\prime }(t)$.
   • D
     $t=6$ only

     Why not D: Used a sign error when factoring $v(t)$.
   Explanation

   Velocity: $v(t)=s^{\prime }(t)=3t^2-18t+24=3(t^2-6t+8)=3(t-2)(t-4)$. Setting $v(t)=0$: $t=2$ or $t=4$. The particle is at rest at these two times.

   Key takeaway

   A particle is at rest when its velocity equals zero; find the zeros of $v(t) = s'(t)$.

9. Question 9 · Medium

   A spherical balloon is being inflated so that its volume increases at $100$ cm³/s. How fast is the radius increasing when the radius is $5$ cm? (Recall $V=\frac{4}{3}\pi r^3$.)

   • A
     $\frac{1}{\pi }$ cm/sCorrect
   • B
     $4\pi$ cm/s

     Why not B: Used $dV/dt=4\pi r^2$ but did not isolate $dr/dt$.
   • C
     $\frac{1}{4\pi }$ cm/s

     Why not C: Forgot to square $r=5$ in the denominator.
   • D
     $\frac{100}{4\pi }$ cm/s

     Why not D: Evaluated at $r=1$ rather than $r=5$.
   Explanation

   $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. At $r=5$, $\frac{dV}{dt}=100$: $100=4\pi (25)\frac{dr}{dt}=100\pi \frac{dr}{dt}$. So $\frac{dr}{dt}=\frac{100}{100\pi }=\frac{1}{\pi }$ cm/s.

   Key takeaway

   Related rates for spheres: differentiate $V = \tfrac{4}{3}\pi r^3$ to get $dV/dt = 4\pi r^2 (dr/dt)$, then solve.

10. Question 10 · Hard

    A 5-foot tall person walks away from a 20-foot lamppost at $3$ ft/s. How fast is their shadow lengthening?

    • A
      $1$ ft/sCorrect
    • B
      $3$ ft/s

      Why not B: Confused the rate of the shadow tip with the rate of shadow length.
    • C
      $4$ ft/s

      Why not C: Found the rate at which the shadow tip moves, not the shadow length.
    • D
      $\frac{3}{4}$ ft/s

      Why not D: Used the ratio $5/20$ directly without the correct similar-triangle setup.
    Explanation

    Let $x$ = distance of person from post, $s$ = shadow length. Similar triangles: $\frac{20}{x+s}=\frac{5}{s}$, so $20s=5(x+s)\Rightarrow 15s=5x\Rightarrow s=\frac{x}{3}$. Differentiate: $\frac{ds}{dt}=\frac{1}{3}\frac{dx}{dt}=\frac{1}{3}(3)=1$ ft/s.

    Key takeaway

    Shadow problems use similar triangles to relate shadow length to person's distance; differentiate that algebraic relation.

11. Question 11 · Hard

    Use L'Hôpital's rule to find $\underset{x\to 0^{+}}{\lim }x\ln x$.

    • A
      $1$

      Why not A: Misidentified the limit as $e^0$.
    • B
      $-1$

      Why not B: Sign or algebra error when applying L'Hôpital.
    • C
      $0$Correct
    • D
      $-\infty$

      Why not D: Noted that $\ln x\to -\infty$ without accounting for $x\to 0^{+}$ pulling the product to zero.
    Explanation

    Rewrite as $\frac{\ln x}{1/x}$ ($\infty /\infty$ form). Apply L'Hôpital: $\frac{1/x}{-1/x^2}=\frac{1/x\cdot x^2}{-1}=-x\to 0$ as $x\to 0^{+}$.

    Key takeaway

    Indeterminate $0 \cdot \infty$ forms are converted to $\infty/\infty$ or $0/0$ before applying L'Hôpital.

12. Question 12 · Hard

    Water drains from a conical tank (vertex down) of height 10 m and radius 4 m at $2$ m³/min. How fast is the water level dropping when $h=5$ m? (Recall $V=\frac{1}{3}\pi r^{2}h$.)

    • A
      $-\frac{1}{2\pi }$ m/minCorrect
    • B
      $-\frac{2}{25\pi }$ m/min

      Why not B: Forgot to evaluate $h^2=25$ and compute the coefficient correctly.
    • C
      $-\frac{1}{4\pi }$ m/min

      Why not C: Used the base radius 4 instead of the similar-triangle radius $r=2h/5$.
    • D
      $-\frac{5}{2\pi }$ m/min

      Why not D: Differentiated without applying the chain rule to $h^3$, dropping a factor of $3$.
    Explanation

    Similar triangles: $r/h=4/10$, so $r=2h/5$. Substituting into volume: $V=\frac{1}{3}\pi {\left(\frac{2h}{5}\right)}^{2}h=\frac{4\pi }{75}{h}^3$. Differentiate: $\frac{dV}{dt}=\frac{4\pi h^2}{25}\frac{dh}{dt}$. At $h=5$ with $\frac{dV}{dt}=-2$: $-2=\frac{4\pi (25)}{25}\frac{dh}{dt}=4\pi \frac{dh}{dt}$. So $\frac{dh}{dt}=-\frac{2}{4\pi }=-\frac{1}{2\pi }$ m/min.

    Key takeaway

    Conical tank related rates: express $r$ in terms of $h$ via similar triangles, substitute into $V = \tfrac{1}{3}\pi r^2 h$, then differentiate.