# Thread: Rate of a decreasing spherical balloon

1. ## Rate of a decreasing spherical balloon

I know:
• Formula for a sphere's volume: $\displaystyle V = \frac{4}{3}*\pi*r^3$
• Diameter of balloon is 20cm = radius of balloon is 10cm.
• $\displaystyle \frac{4}{3}*\pi*10^3 = 1333.3*\pi$
• Difference of 5cm in the radius is: $\displaystyle \frac{4}{3}*pi*5^3 = 166.67*\pi$
• 1333.3 - 166.67 = 1166.63
• This is not a choice, so I am obviously wrong.
• Why I think I am wrong: $\displaystyle 1166.63*\pi cm^3$ is not in the form of a rate. It needs to be in the form of $\displaystyle \frac{x cm^3}{y seconds}$.
• I can't think of how to do this.
• Lastly, my practice test states that this question will see if I understand the historical development and application of calculus. I clearly did not use Calculus, and quite honestly, I'm don't know how.

2. Originally Posted by EMyk01

I know:
• Formula for a sphere's volume: $\displaystyle V = \frac{4}{3}*\pi*r^3$
• Diameter of balloon is 20cm = radius of balloon is 10cm.
• $\displaystyle \frac{4}{3}*\pi*10^3 = 1333.3*\pi$
• Difference of 5cm in the radius is: $\displaystyle \frac{4}{3}*pi*5^3 = 166.67*\pi$
• 1333.3 - 166.67 = 1166.63
• This is not a choice, so I am obviously wrong.
• Why I think I am wrong: $\displaystyle 1166.63*\pi cm^3$ is not in the form of a rate. It needs to be in the form of $\displaystyle \frac{x cm^3}{y seconds}$.
• I can't think of how to do this.
• Lastly, my practice test states that this question will see if I understand the historical development and application of calculus. I clearly did not use Calculus, and quite honestly, I'm don't know how.
1. The rate of change of the volume wrt the time t is:

$\displaystyle \dfrac{dV}{dt}=\dfrac{d\left(\frac43 \pi r^3\right)}{dt}$

2. The rate of change of the radius is $\displaystyle \dfrac{dr}{dt} = 5\ \frac{cm^3}{s}$

3. Calculate dV using the equation from 1. and the value of 2.