172,898 views
14 votes
14 votes
If a snowball melts so that its surface area decreases at a rate of 4 cm²/min, find the rate at which the diameter

decreases when the diameter is 39 cm.

User Nantoka
by
2.9k points

1 Answer

9 votes
9 votes

The surface area of a snowball can be represented by the formula:

A = 4πr^2

where A is the surface area, r is the radius of the snowball, and π is a constant approximately equal to 3.14.

If the surface area of the snowball decreases at a rate of 4 cm²/min, we can represent this as a derivative:

dA/dt = -4

where dA/dt is the rate of change of the surface area with respect to time, and t is the time in minutes.

The radius of the snowball is equal to half the diameter, so we can represent the radius as:

r = d/2

where d is the diameter of the snowball.

Substituting this expression for r into the formula for the surface area and differentiating with respect to time, we get:

dA/dt = 8π(d/2)dr/dt

Setting this expression equal to -4 and solving for dr/dt, we get:

dr/dt = -4 / (8π)

= -0.5 / π

When the diameter of the snowball is 39 cm, the rate at which the diameter decreases is:

dr/dt = -0.5 / π

= -0.15915494309189533576888376337251

Therefore, the rate at which the diameter of the snowball decreases when the diameter is 39 cm is approximately -0.159 cm/min.

I hope this helps

User Miro Kropacek
by
3.2k points