## anonymous 5 years ago If a snowball melts so that its surface area decreases at a rate of , find the rate at which the diameter decreases when the diameter is

1. bahrom7893

carra im on it

2. bahrom7893

what are the rates. Are any numbers given?

3. anonymous

sorry...if a snowball melts so that its surface area decreases at a rate of 0.3cm^2/min. find the rate at which the diameter decreases when the diameter is 15cm.

4. anonymous

i thought it would be dA/dt=0.3cm^2...so would it be dD/dt=2*pi*(7.5)*0.3

5. bahrom7893

sorry will respond in a few mins, i have to do something

6. bahrom7893

Okay so you have: $A = 4\pi r ^{2}$ <=Surface area of a ball

7. bahrom7893

Now you know that d = 2r (diameter is twice the radius), therefore r = d/2 (radius is half of the diameter) Rewrite the equation in terms of diamater

8. bahrom7893

$A = 4\pi (d/2)^{2} = 4\pi * (d ^{2}/2^{2}) = 4\pi * (d^{2}/4)$

9. bahrom7893

Sorry meant diameter in the previous reply. Now you know that: A = 4 pi * (d^2/4), cancel the 4s to get: $A = \pi * d ^{2}$

10. bahrom7893

Now differentiate with respect to t, or time: $dA/dt = (2 *\pi * d) * dd/dt$

11. bahrom7893

You know the change in surface area; and the diameter at which you are required to find the change in radius: dA/dt = -0.3 cm^2/min (THE "-" SIGN HAS TO BE THERE BECAUSE THE SURFACE AREA IS DECREASING) and the diameter = 15 cm so plug those into your equation

12. bahrom7893

$-0.3 = 2\pi*15*(dd/dt)$

13. bahrom7893

-0.3 = 30pi * (dd/dt) [ -0.3 / (30pi) ] = dd/dt $dd/dt = -0.3 / (30\pi)$

14. bahrom7893

woops forgot units, just add cm/min at the end, check with the answer

15. bahrom7893

sorry carra wrong answer, lucky someone caught me

16. bahrom7893

i was right

17. anonymous

then y am i getting dd/dt=2*pi*15*0.3

18. bahrom7893

Okay let me try this again; I will write it out on a piece of paper and upload it, i hate typing

19. bahrom7893

http://i55.tinypic.com/8wdsp1.jpg anything unclear? there? ask me, i can explain steps