If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm.

Question

asked Aug 10, 2022 201k views

1 Answer

Vpap · Answer 1 · 2022-08-17T17:31:05+0000

Answer:

The diameter decreases at a rate of -0.0477 cm/min when it is 10 cm.

Explanation:

Surface area of a snowball:

A snowball has a spheric format, which means that it's surface area is given by:

$A = 4\pi r^2$

In which r is the radius, which is half the diameter. In function of the diameter, the area is given by:

$A = 4\pi((d)/(2))^2 = \pi d^2$

Solving the question:

To solve this question, we have to implicitly derivate the area in function of t. So

$(dA)/(dt) = 2d\pi(dd)/(dt)$

Snowball melts so that its surface area decreases at a rate of 3 cm2/min

This means that
(dA)/(dt) = -3

Find the rate at which the diameter decreases when the diameter is 10 cm.

This is
(dd)/(dt) when
d = 10 . So

$(dA)/(dt) = 2d\pi(dd)/(dt)$

$-3 = 20\pi(dd)/(dt)$

$(dd)/(dt) = -(3)/(20\pi)$

(dd)/(dt) = -0.0477

The diameter decreases at a rate of -0.0477 cm/min when it is 10 cm.