Rate of change of cross-section of cylinder 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 dD/dt=0,2cm/sec

Kade Reese

Kade Reese

Answered question

2022-07-14

Rate of change of cross-section of cylinder
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
d D d t = 0 , 2 cm / sec
Rate of change of height is
d h d t = 0 , 4 cm / sec
Trying to find d D d t by using formula
V = π r 2 h

Answer & Explanation

yelashwag8

yelashwag8

Beginner2022-07-15Added 17 answers

So, First we know that
V = π D 2 h 4
where D is the Diameter, and h is the height. Now by differentiating this with respect to time, we get,
d V d t = V D d D d t   + V h d h d t
= π D h 2 d D d t   + π D 2 4 d h d t
Now, d V d t = 0 ; d D d t = 2 ; d h d t = 4 ; a n d h = 10
(Note: All units are in mm)
Solving for D we get, D=10mm
Now coming back to the question we are needed to find
d A d t
where A = π D 2 4 (Cross sectional area)
Taking derivative with respect to time, we get, d A d t = π D 2 d D d t
Plugging in values, we obtain,
d A d t = 10 π m m 2 / s

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?