Related Rates Practice

Sample questions

5 of 20 — sign in to practice the rest with adaptive difficulty and mastery tracking.

1. Sample 1difficulty 2/5

   A square's side grows at $3~\text{cm/s}$. When the side is $4~\text{cm}$, its area grows at

   • A

     $24~\text{cm}^2/\text{s}$

     check_circle
   • B

     $12~\text{cm}^2/\text{s}$

   • C

     $3~\text{cm}^2/\text{s}$

   • D

     $48~\text{cm}^2/\text{s}$

   Why

   $A = s^2$, $dA/dt = 2s\,ds/dt = 2(4)(3) = 24~\text{cm}^2/\text{s}$.

   View full solution & statsarrow_forward

2. Sample 2difficulty 2/5

   A rectangle's length grows at $2$ and width shrinks at $1$ (units/s). When $L = 5, W = 3$, the area changes at

   • A

     $3$

   • B

     $1$

     check_circle
   • C

     $6$

   • D

     $11$

   Why

   $A = LW$, $dA/dt = L W' + L' W = 5(-1) + 2(3) = 1$.

   View full solution & statsarrow_forward

3. Sample 3difficulty 2/5

   A cube's edge length grows at $2~\text{cm/s}$. When edge is $5~\text{cm}$, its volume grows at

   • A

     $150~\text{cm}^3/\text{s}$

     check_circle
   • B

     $30~\text{cm}^3/\text{s}$

   • C

     $250~\text{cm}^3/\text{s}$

   • D

     $50~\text{cm}^3/\text{s}$

   Why

   $V = s^3$, $dV/dt = 3s^2\,ds/dt = 3(25)(2) = 150~\text{cm}^3/\text{s}$.

   View full solution & statsarrow_forward

4. Sample 4difficulty 3/5

   A snowball's radius melts at $-0.5~\text{cm/min}$. When $r = 3~\text{cm}$, the surface area changes at (use $S = 4\pi r^2$):

   • A

     $-24\pi~\text{cm}^2/\text{min}$

   • B

     $-3\pi~\text{cm}^2/\text{min}$

   • C

     $-12\pi~\text{cm}^2/\text{min}$

     check_circle
   • D

     $-36\pi~\text{cm}^2/\text{min}$

   Why

   $dS/dt = 8\pi r\,dr/dt = 8\pi(3)(-0.5) = -12\pi$.

   View full solution & statsarrow_forward

5. Sample 5difficulty 3/5

   A balloon rises straight up at $5~\text{ft/s}$. An observer is $100~\text{ft}$ from the launch point. When the balloon is $100~\text{ft}$ high, the line of sight changes angle (in radians) at

   • A

     $\tfrac{1}{200}~\text{rad/s}$

   • B

     $\tfrac{1}{50}~\text{rad/s}$

   • C

     $\tfrac{1}{20}~\text{rad/s}$

   • D

     $\tfrac{1}{40}~\text{rad/s}$

     check_circle

   Why

   $\tan\theta = h/100$, $\sec^2\theta\,d\theta/dt = (1/100)dh/dt$. At $h = 100$, $\theta = 45^\circ$, $\sec^2 = 2$. So $d\theta/dt = (1/100)(5)/2 = 1/40~\text{rad/s}$.

   View full solution & statsarrow_forward