Question

Calculus Related Rates:
If a snowball melts so that its surface area decreases at a rate of 7 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm.
Please show all of your work... and tell me why I keep getting -.0139?

Answer 1

SA=4pi r^2 or 4pi (1/2D)^2

Answer 2

\[S(D)= \pi D^2\] so
\[S'=2\pi D D'\]

Answer 3

you are told that
\[S'=-7\] so solve for
\[D'\] and you have it

Answer 4

A =Surface Area of a Sphere = 4pi*r^2
dA/dt = - 7cm^2/min

dA/dt = dA/dr * dr/dt, where dr/dt is change in radius wrt time

dA/dt = 8 * pi *r * dr/dt = -7 cm^2/min

so dr/dt = -7/(pi*r*8)
= -7/pi*5*8)
= -.056 cm/min

Ans. -.056 cm/min

Answer 5

dA / dt = dA/ dD * dD/dt
-7 = 2*pi* D * dD / dt

Answer 6

\[SA=4 \pi r^2 => SA'=4 \pi (2 r r')=8 \pi r r' \]
where SA is the surface area
\[SA'=-7\]
\[d=10 => r=\frac{10}{2}=5\]
we need to find d'
d=2r => d'=2r'

so we have
\[-7=8 \pi (5) r' => r'=\frac{-7}{8 \pi (5)}=\frac{-7}{40 \pi} => d'=2r'=2(\frac{-7}{40 \pi})=\frac{-7}{20 \pi}\]

Answer 7

ybarrap- Your answer would be in terms of r right, not D?
And give me a second to work this through now, and I will let you all know if Webassign agrees.

Answer 8

Okay thanks everyone. The answer was actually positive because it asked for the rate in which it decreases. It was 7/20pi

Answer 9

i get rate of change = -7/20 pi - same as myin - thatgs rate of change of diameter at d = 10 cm