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 8 cm.

Question

baru · Answer

http://openstudy.com/users/baru#/updates/563719dbe4b027f3ddc2798a

campbell_st · Answer

so this is a related rates question 

so is the snowball a sphere...

baru · Answer

answered same ques ^

anonymous · Answer

Yes it is a sphere campbell_st

campbell_st · Answer

ok... so using the diameter in stead of the radius  in the surface area of a sphere it becomes

$$S = \pi \times (\frac{d}{2})^2 ~~or~~ S = \frac{ \pi \times d^2}{4}$$

does that make sense...?

anonymous · Answer

Oh! thanks  baru...It's not exactly the same but I'll see

anonymous · Answer

Shouldn't it be 4pi(d/2)^2 ? Campbell_st

campbell_st · Answer

oops forgot the 4 $$S = 4 \pi (\frac{d}{2})^2 ~~~or~~S = \frac{4\pi d^2}{4} $$

so the surface area in terms of diameter is 

$$S = \pi d^2$$

sorry about the error

campbell_st · Answer

so can you find the derivative od S with respect to d...?

anonymous · Answer

That will be 2pi*d right?

campbell_st · Answer

correct

so you now know 

$$\frac{dS}{dD} = 2\pi d$$

you also know 

$$\frac{dS}{dt} = 10 $$

you could use -10... but everything is decreasing... 

so you need to use 

$$\frac{dD}{dt} = \frac{dD}{dS} \times \frac{dS}{dt}$$

so find the reciprocal of $$\frac{dS}{dD} ~~then~~multiply~~by~~\frac{dS}{dt}$$

to get the equation forthe  rate of change in the diameter with respect to time... 

then just substitute d = 8 for the answer... 

hope it helps

anonymous · Answer

So then the final answer would be 5/8pi ?

campbell_st · Answer

That's what I got... and it should be cm/min

anonymous · Answer

Thank you so much Campbell!!! :-D

campbell_st · Answer

glad to help