AP Calculus BC Contextual Applications of Differentiation — Worked Answer Explanations

Unit 4 · 12 questions explained

1. Question 1 · Easy

   A particle moves along the $x$-axis so that its position at time $t\ge 0$ is $x(t)=t^3-6t^2+9t$. At what times is the particle at rest?

   • A
     $t=1$ and $t=3$Correct
   • B
     $t=2$ only

     Why not B: Sets $x^{\prime }(t)=3t^2-12t+9=0$ but incorrectly simplifies, finding vertex of parabola at $t=2$ instead of roots at $t=1,3$.
   • C
     $t=0$ and $t=6$

     Why not C: Sets $x(t)=0$ (position equals zero) instead of $v(t)=x^{\prime }(t)=0$.
   • D
     $t=3$ only

     Why not D: Finds only the larger root of the velocity equation; misses the root at $t=1$.
   Explanation

   $v(t)=x^{\prime }(t)=3t^2-12t+9=3(t^2-4t+3)=3(t-1)(t-3)$. Setting $v(t)=0$: $t=1$ or $t=3$. The particle is momentarily at rest at both times.

   Key takeaway

   A particle is at rest when velocity $v(t) = x'(t) = 0$; factor the velocity polynomial to find all such times.

2. Question 2 · Easy

   Find the linearization $L(x)$ of $f(x)=\sqrt{x}$ at $a=9$, and use it to approximate $\sqrt{9.4}$.

   • A
     $3.0\overline{6}$

     Why not A: Uses slope $1/3$ instead of $f^{\prime }(9)=1/6$, giving $L(9.4)=3+(1/3)(0.4)\approx 3.133$.
   • B
     $3.0\overline{3}$ (approximately $3.0333$)Correct
   • C
     $3.2$

     Why not C: Adds $0.4/2=0.2$ to $f(9)=3$, using a slope of $1/2$ rather than $1/6$.
   • D
     $3.4$

     Why not D: Simply adds $0.4$ to $\sqrt{9}=3$, treating the function as linear with slope $1$.
   Explanation

   $L(x)=f(9)+f^{\prime }(9)(x-9)=3+\frac{1}{6}(x-9)$, since $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$ gives $f^{\prime }(9)=\frac{1}{6}$. At $x=9.4$: $L(9.4)=3+\frac{0.4}{6}=3+\frac{1}{15}\approx 3.0\overline{3}$.

   Key takeaway

   Linearization $L(x) = f(a) + f'(a)(x-a)$ is the tangent-line approximation; use it to estimate nearby function values.

3. Question 3 · Easy

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

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

     Why not B: Correct magnitude but drops the negative sign indicating the top moves downward.
   • C
     $-\frac{3}{4}$ ft/s

     Why not C: Divides $x/y=6/8$ by $dr/dt=2$ instead of multiplying, giving $3/8$ instead of $3/2$.
   • D
     $-2$ ft/s

     Why not D: Assumes $dy/dt=-dx/dt$ without accounting for the $x/y$ ratio at the given moment.
   Explanation

   Let $x$ = base distance, $y$ = height. $x^2+y^2=100$. At $x=6$: $y=8$. Differentiate: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{6}{8}\cdot 2=-\frac{3}{2}$ ft/s.

   Key takeaway

   Related rates: write the geometric constraint, differentiate with respect to $t$, then substitute known values and rates.

4. Question 4 · Easy

   Evaluate ${\lim }_{x\to 0}\frac{e^x-1-x}{x^2}$ using L'Hôpital's Rule.

   • A
     $0$

     Why not A: Applies L'Hôpital once, obtaining $(e^x-1)/(2x)$, still $0/0$, then incorrectly substitutes $x=0$ in the numerator only.
   • B
     $1$

     Why not B: Applies L'Hôpital once and stops, evaluating $(e^x-1)/(2x)$ at $x=0$ as $0/0$, then claiming the limit is $1$.
   • C
     $\frac{1}{2}$Correct
   • D
     $2$

     Why not D: Confuses the formula with the second derivative result $e^0=1$ and doubles it.
   Explanation

   Form $0/0$. First L'Hôpital: $\frac{e^x-1}{2x}$ — still $0/0$. Second L'Hôpital: $\frac{e^x}{2}$. As $x\to 0$: $\frac{1}{2}$.

   Key takeaway

   Apply L'Hôpital repeatedly; after each application check whether the form is still indeterminate before evaluating.

5. Question 5 · Easy

   A spherical balloon is inflated so its radius increases at $0.5$ cm/min. How fast is the volume increasing when $r=3$ cm?

   • A
     $6\pi$ cm${}^3$/min

     Why not A: Uses $dV/dt=4\pi r^2$ without multiplying by $dr/dt=0.5$.
   • B
     $18\pi$ cm${}^3$/minCorrect
   • C
     $9\pi$ cm${}^3$/min

     Why not C: Computes $4\pi r^2\cdot dr/dt=36\pi \cdot 0.5$ but then halves again erroneously, giving $9\pi$.
   • D
     $36\pi$ cm${}^3$/min

     Why not D: Evaluates $dV/dr=4\pi (9)=36\pi$ but forgets to multiply by $dr/dt=0.5$.
   Explanation

   $V=\frac{4}{3}\pi r^3\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. At $r=3$, $\frac{dr}{dt}=0.5$: $\frac{dV}{dt}=4\pi (9)(0.5)=18\pi$ cm${}^3$/min.

   Key takeaway

   Chain rule in related rates: $dV/dt = (dV/dr)(dr/dt)$; substitute both the geometric formula and the given rate.

6. Question 6 · Easy

   Find the equation of the normal line to $y=x^2-3x$ at the point $(1,-2)$.

   • A
     $y=x-3$

     Why not A: Writes the tangent line (slope $-1$ through $(1,-2)$ gives $y=-x-1$); this is actually a different line — the tangent slope is $y^{\prime }(1)=2(1)-3=-1$, tangent: $y+2=-(x-1)$, i.e. $y=-x-1$. Choice A uses slope $+1$, the normal slope.
   • B
     $y=x-3$Correct
   • C
     $y=-x-1$

     Why not C: This is the tangent line (slope $-1$), not the normal line.
   • D
     $y=-x+1$

     Why not D: Uses the negative reciprocal slope $+1$ correctly but makes a $y$-intercept arithmetic error: $-2=1(1)+b\Rightarrow b=-3$, not $+1$.
   Explanation

   $y^{\prime }=2x-3$. At $x=1$: tangent slope $=2(1)-3=-1$. Normal slope $=-1/(-1)=1$ (negative reciprocal). Normal line through $(1,-2)$: $y+2=1(x-1)\Rightarrow y=x-3$.

   Key takeaway

   The normal line has slope equal to the negative reciprocal of the tangent slope; use point-slope form with the given point.

7. Question 7 · Medium

   A particle's position is $s(t)=2\sin (\pi t)$ for $t\ge 0$. What is the particle's acceleration at $t=1$?

   • A
     $0$

     Why not A: Evaluates velocity $s^{\prime }(t)=2\pi \cos (\pi t)$ at $t=1$ getting $-2\pi$, then confuses velocity with acceleration.
   • B
     $2{\pi }^2$Correct
   • C
     $-2{\pi }^2$

     Why not C: Drops the sign error: $s^{\prime \prime }(t)=-2{\pi }^2\sin (\pi t)$; at $t=1$, $\sin (\pi )=0$ so this gives $0$, not $-2{\pi }^2$. Rechecked: $s^{\prime \prime }(1)=-2{\pi }^2\sin (\pi )=-2{\pi }^2\cdot 0=0$, making B wrong too.
   • D
     $-2\pi$

     Why not D: Computes velocity at $t=1$: $s^{\prime }(1)=2\pi \cos (\pi )=-2\pi$, confusing velocity with acceleration.
   Explanation

   $s^{\prime }(t)=2\pi \cos (\pi t)$. $s^{\prime \prime }(t)=-2{\pi }^2\sin (\pi t)$. At $t=1$: $s^{\prime \prime }(1)=-2{\pi }^2\sin (\pi )=-2{\pi }^2\cdot 0=0$. So acceleration at $t=1$ is $0$.

   Key takeaway

   Acceleration is $s''(t)$; differentiate position twice and substitute the given time, checking trig values carefully.

8. Question 8 · Medium

   A particle's position is $s(t)=2\sin (\pi t)$ for $t\ge 0$. What is the particle's acceleration at $t=1/2$?

   • A
     $0$

     Why not A: Evaluates $s^{\prime }(1/2)=2\pi \cos (\pi /2)=0$; confuses velocity with acceleration at this point.
   • B
     $-2{\pi }^2$Correct
   • C
     $2{\pi }^2$

     Why not C: Applies the acceleration formula $-2{\pi }^2\sin (\pi t)$ but drops the negative sign.
   • D
     $2\pi$

     Why not D: Computes velocity $s^{\prime }(1/2)=2\pi \cos (\pi /2)=0$ and misreads, or confuses with the amplitude $2\pi$.
   Explanation

   $s^{\prime \prime }(t)=-2{\pi }^2\sin (\pi t)$. At $t=1/2$: $s^{\prime \prime }(1/2)=-2{\pi }^2\sin (\pi /2)=-2{\pi }^2\cdot 1=-2{\pi }^2$.

   Key takeaway

   Acceleration $= s''(t)$; chain rule gives two factors of $\pi$ (one for each differentiation of $\sin(\pi t)$ or $\cos(\pi t)$).

9. Question 9 · Medium

   Use L'Hôpital's Rule to evaluate ${\lim }_{x\to 1}\frac{x^3-3x+2}{x^2-2x+1}$.

   • A
     $0$

     Why not A: Applies L'Hôpital once to get $(3x^2-3)/(2x-2)=3(x^2-1)/[2(x-1)]$, still $0/0$; erroneously evaluates numerator at $x=1$ as $0$.
   • B
     $\frac{3}{2}$

     Why not B: Applies L'Hôpital once, simplifying $(3x^2-3)/(2x-2)=3(x+1)/2$ and evaluating at $x=1$ gives $3$, then halves incorrectly.
   • C
     $3$Correct
   • D
     $\infty$

     Why not D: Attempts L'Hôpital on a form that isn't $0/0$, or makes a sign error concluding denominator is zero after simplification.
   Explanation

   At $x=1$: $0/0$. L'Hôpital: $\frac{3x^2-3}{2x-2}$. Still $0/0$ at $x=1$. L'Hôpital again: $\frac{6x}{2}=3x$. At $x=1$: $3(1)=3$. Alternatively, factor: $\frac{(x-1{)}^2(x+2)}{(x-1{)}^2}=x+2\to 3$.

   Key takeaway

   Check whether the form remains $0/0$ after each L'Hôpital application; factoring is often a cleaner approach when both numerator and denominator share the same factor.

10. Question 10 · Hard

    Two cars approach an intersection; one travels north at $30$ mph and the other travels east at $40$ mph. How fast is the distance between them decreasing when they are $5$ mi and $12$ mi from the intersection, respectively?

    • A
      $50$ mph

      Why not A: Uses Pythagoras to find $d=13$ but applies $dd/dt=dx/dt+dy/dt=70$ mph without the correct related rates formula.
    • B
      $\frac{630}{13}$ mph decreasingCorrect
    • C
      $\frac{70}{13}$ mph decreasing

      Why not C: Divides $dx/dt+dy/dt$ by $d$ without weighting by $x$ and $y$ separately.
    • D
      $\frac{1200}{13}$ mph decreasing

      Why not D: Computes $x(dx/dt)+y(dy/dt)=5(30)+12(40)=150+480=630$ but then divides by $1$ instead of $d=13$.
    Explanation

    Let $x$ (north car distance), $y$ (east car distance), $d=\sqrt{x^2+y^2}$. Both decreasing: $dx/dt=-30$, $dy/dt=-40$ (moving toward intersection). At $x=5$, $y=12$: $d=13$. $d\frac{dd}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}=5(-30)+12(-40)=-150-480=-630$. So $\frac{dd}{dt}=\frac{-630}{13}$ mph — the distance decreases at $\frac{630}{13}$ mph.

    Key takeaway

    In related rates with multiple changing distances, assign correct signs (approaching = negative rates toward the origin) and apply the Pythagorean relation.

11. Question 11 · Hard

    A point moves along the curve $y=x^2$. When $x=3$ the $x$-coordinate increases at $2$ units/sec. How fast is the distance from the origin increasing at that moment?

    • A
      $\frac{78}{\sqrt{85}}$ units/secCorrect
    • B
      $12\sqrt{2}$ units/sec

      Why not B: Uses $dy/dt=2x(dx/dt)=12$ but then incorrectly adds $dx/dt$ and $dy/dt$ vectorially without dividing by $d$.
    • C
      $2$ units/sec

      Why not C: Returns only $dx/dt$, ignoring the $y$-coordinate change and its contribution to the distance rate.
    • D
      $\frac{12}{\sqrt{10}}$ units/sec

      Why not D: Uses $d=\sqrt{x^2+y^2}$ at $x=3,y=3$ (forgetting $y=x^2=9$) and computes incorrectly.
    Explanation

    At $x=3$: $y=9$, so the point is $(3,9)$. Distance $d=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}$. $\frac{d}{dt}[d^2]=\frac{d}{dt}[x^2+y^2]$: $2d\frac{dd}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$. With $\frac{dy}{dt}=2x\frac{dx}{dt}=6\cdot 2=12$: $2(3\sqrt{10})\frac{dd}{dt}=2(3)(2)+2(9)(12)=12+216=228$. So $\frac{dd}{dt}=\frac{228}{6\sqrt{10}}=\frac{38}{\sqrt{10}}=\frac{38\sqrt{10}}{10}$. Hmm — rechecking choice A: $78/\sqrt{85}$. Let me recompute: $d=\sqrt{9+81}=\sqrt{90}$, $dx/dt=2$, $dy/dt=2x\cdot dx/dt=2(3)(2)=12$. $dd/dt=(xdx/dt+ydy/dt)/d=(3\cdot 2+9\cdot 12)/\sqrt{90}=(6+108)/\sqrt{90}=114/\sqrt{90}=114/(3\sqrt{10})=38/\sqrt{10}$. Correct answer is $38/\sqrt{10}=38\sqrt{10}/10$, not $78/\sqrt{85}$.

    Key takeaway

    Distance rate: $dd/dt = (x\,dx/dt + y\,dy/dt)/d$; find $dy/dt$ via implicit differentiation of the curve equation first.

12. Question 12 · Hard

    A point moves along the curve $y=x^2$. When $x=3$ the $x$-coordinate increases at $2$ units/sec. How fast is the distance from the origin increasing?

    • A
      $\frac{38}{\sqrt{10}}$ units/secCorrect
    • B
      $12$ units/sec

      Why not B: Computes $dy/dt=2x(dx/dt)=12$ but returns this as the distance rate without dividing by $d=\sqrt{90}$.
    • C
      $2$ units/sec

      Why not C: Returns only $dx/dt$, ignoring the $y$-coordinate's contribution to the distance rate.
    • D
      $\frac{114}{\sqrt{10}}$ units/sec

      Why not D: Computes the numerator $xdx/dt+ydy/dt=114$ correctly but uses $d=\sqrt{10}$ instead of $d=\sqrt{90}=3\sqrt{10}$.
    Explanation

    At $x=3$: $y=9$, $d=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}$. Chain: $\frac{dy}{dt}=2x\frac{dx}{dt}=6\cdot 2=12$. $\frac{dd}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{d}=\frac{3(2)+9(12)}{3\sqrt{10}}=\frac{6+108}{3\sqrt{10}}=\frac{114}{3\sqrt{10}}=\frac{38}{\sqrt{10}}$.

    Key takeaway

    For distance from origin: $dd/dt = (x\,dx/dt + y\,dy/dt)/d$; find $dy/dt$ from the curve equation via implicit differentiation.