# Related Rates [Balloon]

• Mar 16th 2009, 06:29 PM
Raj
Related Rates [Balloon]
The radius of a speherical balloon is decreasing at a rate of 3.5 cm/s. At what rate is the surface area of the balloon decreasing when the radius is 14cm?

dr/dt = -3.5 cm/s

da/dt = ? when r = 14

Ive used the area of a sphere forumla, derived, plugged in numbers but i'm getting a wrong answer.
• Mar 16th 2009, 06:37 PM
Reckoner
Quote:

Originally Posted by Raj
Ive used the area of a sphere forumla, derived, plugged in numbers but i'm getting a wrong answer.

Could you show us your work? We can point out your mistake.
• Mar 16th 2009, 06:39 PM
Quote:

Originally Posted by Raj
The radius of a speherical balloon is decreasing at a rate of 3.5 cm/s. At what rate is the surface area of the balloon decreasing when the radius is 14cm?

dr/dt = -3.5 cm/s

da/dt = ? when r = 14

Ive used the area of a sphere forumla, derived, plugged in numbers but i'm getting a wrong answer.

Your set up is correct .

$\displaystyle \frac{dA}{dr}=4\pi r^2$

$\displaystyle =8\pi r=112\pi$

just multiply it with -3.5 to get the answer .