If a snowball melts so that its surface area decreases at a rate of 10 cm2/min, find the rate at which the diameter decreases when the diameter is 9 cm. cm/min

4/9π

Explanation:

Surfacearea of snowball(SA) =4πr^2

dSA/dT = - 10cm^2/min

d = 9, hence, r = 9/2 = 4.5

Taking the derivative of surface area with respect to time:

dSA/dT = 8πrdr/dt

dSA/dT = - 10, r = 4.5

-10 = 8π(4.5)dr/dt

-10 = 36πdr/dt

Make dr/dt subject of the formula

dr/dt = - 10/36π

d = 2r

Differentiating d with respect to time, 't'

dd/dt = 2dr/dt

Therefore, 2dr/dt = 2*(dr/dt)

2×(- 10/36π) = - 20/36π

= -4/9π

Therefore, rate by which diameter decreases when diameter is 9cm is

4/9 π

a. vertically

explanation:

i tried it and took the test

if im wrong then we have different tests