I got this task on my calculus class and I got stuck at process of figuring it out
A clay cylinder is being compressed so that its height is changing at the rate of 4 millimeters per second, and its diameter is increasing at the rate of 2 millimeters per second. Find the rate of change of the area of the horizontal cross-section of the cylinder when its height is 1 centimeter.
What I know:
Volume is unknown and constant
Rate of change of diameter is $$\dfrac {dD}{dt} = 0,2\text{cm}/\text{sec}$$
Rate of change of height is $$\dfrac {dh}{dt} = -0,4\text{cm}/\text{sec}$$
Trying to find $\dfrac{dD}{dt}$ by using formula $$V=\pi r^2h$$
Best Answer
So, First we know that $$V=\frac{\pi D^2 h}{4} $$ where $D$ is the Diameter, and $h$ is the height. Now by differentiating this with respect to time, we get,
$\frac{dV}{dt} = \frac {\partial V}{\partial D}\frac{dD}{dt}\ + \frac {\partial V}{\partial h}\frac{dh}{dt}$
$\;\;\;\;\;= \frac{\pi D h}{2} \frac{dD}{dt}\ + \frac{\pi D^2}{4} \frac{dh}{dt}$
Now, $\frac {dV}{dt} = 0 ; \; \frac{dD}{dt}=2; \; \frac{dh}{dt} = -4; \;\;\; and \;\;\; h=10 $
(Note: All units are in mm)
Solving for $D$ we get, $D=10 mm$
Now coming back to the question we are needed to find $$\frac {dA}{dt}$$
where $A= \frac {\pi D^2}{4}$ (Cross sectional area)
Taking derivative with respect to time, we get, $\frac {dA}{dt} = \frac {\pi D}{2} \frac {dD}{dt}$
Plugging in values, we obtain,
$\mathbf {\frac {dA}{dt} = 10 \pi \; \;mm^2/s}$