Question

If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 9 cm

Answer 1

Answer:

0.0221 cm/min

Step-by-step explanation:

Surface area of a snowball:

An snowball has a spherical format.

The surface area of an sphere is given by:

$A = 4\pi r^2$

In which r is the radius(half the diameter). So in function of the diameter, we have that:

$A = 4\pi (\frac{d}{2})^2 = \pi d^2$

Implicit derivative:

To solve this question, we need to find the implicit derivative of A in function of t.

The variables are A and d, so:

$\frac{dA}{dt} = 8\pi d \frac{dd}{dt}$

Its surface area decreases at a rate of 5 cm2/min

This means that $\frac{dA}{dt} = -5$

Find the rate (in cm/min) at which the diameter decreases when the diameter is 9 cm

This is $\frac{dd}{dt}$ when $d = 9$. So

$\frac{dA}{dt} = 8\pi d \frac{dd}{dt}$

$-5 = 8\pi*9 \frac{dd}{dt}$

$\frac{dd}{dt} = -\frac{5}{72\pi}$

$\frac{dd}{dt} = -0.0221$

This means that the diameter decreases at a rate of 0.0221 cm/min when the diameter is 9 cm.