Solve the problem.
-The radius r and height h of a cylinder are changing with time. At the instant in question, r = 2 cm, h = 6 cm, dr/dt = 0.03 cm/sec and dh/dt = -0.03 cm/sec. At what rate is the cylinder's volume changing at that instant?
A) $2.64 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$
B) $0.75 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$
C) $0.94 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$
D) $1.88 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$

Question

Solve the problem.
-The radius r and height h of a cylinder are changing with time. At the instant in question, r = 2 cm, h = 6 cm, dr/dt = 0.03 cm/sec and dh/dt = -0.03 cm/sec. At what rate is the cylinder's volume changing at that instant? 
A) $2.64 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$
B) $0.75 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$
C) $0.94 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$
D) $1.88 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$

Quizplus · Accepted Answer

The formula for the volume of a cylinder is $V = \pi r ^ { 2 } h$. Taking the derivative with respect to time we have, 
\begin{align*}
\frac { dV } { d t } &= \pi \left( 2 r \frac { d r } { d t } ight) h + \pi r ^ { 2 } \frac { d h } { d t } \
&= 2 \pi (2) (0.03) (6) + \pi (2) ^ { 2 } (-0.03) \
&= 1.88 \mathrm {~cm} ^ { 3 } / \mathrm { sec }
\end{align*}

Solve the Problem 2.64 cm3/sec2.64 \mathrm {~cm} ^ { 3 } / \mathrm { sec }2.64 cm3/sec

Solve the Problem $2.64 \mathrm {~cm} ^ { 3 } / \mathrm { sec }$