A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.1 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 10 cm. (Note the answer is a positive number).
- print Print
- list Cite
Expert Answers
briefcaseTeacher (K-12)
calendarEducator since 2011
write3,185 answers
starTop subjects are Math, Science, and Business
We are given the rate of decrease of the diameter `(dd)/(dt)=.1"cm"/"min" ` . We are asked to find the rate of the decrease of the volume when the diameter is 10cm:
The formula for the volume of a sphere is `V=4/3pir^3 ` .
The derivative is ` (dV)/(dt)=4pir^2(dr)/(dt) `
If the diameter is 10cm, the radius is 5cm. Also, if `(dd)/(dt)=.1 ` then `(dr)/(dt)=.05 `
Then substituting the given values:
`(dV)/(dt)=4pi(5)^2(.05)~~15.71 `
(The entire section contains 2 answers and 240 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- Calculus AB Related RatesA spherical snowball is melting in such a way that its volume is...
- 1 Educator Answer
- If a snowball melts so that its surface area decreases at a rate of 1cm^2/min, find the rate at...
- 1 Educator Answer
- The length l of a rectangle is decreasing at the rate of 2 cm/sec while the width w is increasing...
- 1 Educator Answer
- Related rates problem: The length of a rectangle is increasing at a rate of 8 cm/s and its width...
- 2 Educator Answers
- A conical icicle is melting at a rate of 10.0 cm^3/h. At 10:00 am, the icicle is 30 cm long and 8...
- 2 Educator Answers
You can also use the equation:
V = (1/6)pi d^3
dV/dt = (1/2)pi d^2 dd/dt
Since dd/dt = .1 cm/min, and d = 10 cm, we only have to plug in and find that the answer is 15.7 cm^3/min.
Student Answers