melting sphere of ice (snowball melting at 8 mL/min)

[chargr10]

A snowball melts at a rate of 8mL/min. How fast is is the outer surface area decreasing when the diameter is 20cm.

d=20cm
r=10cm

s(t)=8mL/min=8cm^3/min

SA=4*pi*r^2

SA=4*pi*100 SA=400pi cm^2

V=4/3*pi*r3

V=4/3*pi*(10)^3

so is my next move s(t)= 4*pi*2r*dr/dt
8cm^3/min=4*pi*20cm*dr/dt

8cm^3/min/80pi cm=dr/dt

answer should be 1.6 so I am lost

 

[arthur ohlsten]

Re: melting sphere of ice (snowball)

let V = volume
r=radius [10 cm]
dV/dt = 8 cm^3/min

V=4/3 pi r^3 take derivative with respect to t to find dr/dt
dV/dt = 4 pi r^2 dr/dt
8 cm^3/min = 4 pi [100] dr/dt cm^2
dr/dt = 1/[50pi] cm/min

but we want the change of Area [A]
A= 4pir^2 take derivative
dA/dt = 8 pi r dr/dt
dA/dt =8 pi [10]cm [1/50pi] cm/min
dA/dt = [8/5]cm^2/min
dA/dt= 1.6 cm^2/min answer

now that you see the answer , redo the problem
Arthur

 

[chargr10]

Re: melting sphere of ice (snowball)

thank you. you make it seem so easy

 

[mmm4444bot]

Re: melting sphere of ice (snowball)

... you make it seem so easy

What is "it", exactly?



"Spoon feeding, in the long run, teaches us nothing but the shape of the spoon." ~ E. M. Forster

 

[chargr10]

"it" would be stting up the problem. I am having trouble setting them up properly.