The radius of a sphere increases at a rate

DofotheroU 2021-10-02 Answered
The radius of a sphere increases at a rate of \(\displaystyle{4}\frac{{m}}{{\sec{}}}\). Find the rate at which the surface area increases when the radius is 9m.
\(\displaystyle{288}\pi\frac{{m}^{{{2}}}}{{\sec{}}}\)
\(\displaystyle{36}\pi\frac{{m}^{{{2}}}}{{\sec{}}}\)
\(\displaystyle{18}\pi\frac{{m}^{{{2}}}}{{\sec{}}}\)
\(\displaystyle{72}\pi\frac{{m}^{{{2}}}}{{\sec{}}}\)
\(\displaystyle{144}\pi\frac{{m}^{{{2}}}}{{\sec{}}}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

krolaniaN
Answered 2021-10-03 Author has 12799 answers

Step 1
Surface area of sphere is \(\displaystyle{S}={4}\pi{r}^{{{2}}}\).
And rate of change means derivatives.
Step 2
Now let radius of sphere is =r.
So dr/dt=4.
So rate of change of surface area \(\displaystyle={d}\frac{{S}}{{\left.{d}{t}\right.}}={8}\pi{r}.{}\frac{{dr}}{{\left.{d}{t}\right.}}={8}\pi{.9}{.4}={288}\pi\).
So options A correct.

Have a similar question?
Ask An Expert
29
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-05-04
The radius of a sphere increases at a rate of \(4m/\sec\). Find the rate at which the surface area increases when the radius is 9m.
\(288\pi m^{2}/\sec\)
\(36\pi m^{2}/\sec\)
\(18\pi m^{2}/\sec\)
\(72\pi m^{2}/\sec\)
\(144\pi m^{2}/\sec\)
asked 2021-10-06
A sphere has its radius increasing at a rate of 2 mm/s. How fast is the volume increasing when the diameter is 100 mm?
asked 2021-08-14
The radius of a sphere is increasing at a rate of 2 mm/s. How fast is the volume increasing when the diameter is 100 mm?
asked 2020-12-12

Show that at the bottom of a vertical mine shaft dug to depthD, the measured value of ag will be \(ag = a_{gs}(1 - D/R)\), \(a_{gs}\) being the surface value. Assume that Earth is a uniform sphere of radius R.

asked 2021-03-30
A charged belt, 50 cm wide, travels at 30 m/s between a source ofcharge and a sphere. The belt carries charge into the sphereat a rate corresponding to 100 (micro)amperes. Compute thesurface charge density on the belt.
asked 2021-03-25
Ea for a certain biological reaction is 50 kJ/mol, by what factor ( how many times) will the rate of this reaction increase when body temperature increases from 37 C (normal ) to 40 C (fewer)?
asked 2021-06-14
A wheel with a 30-cm radius is rotating at a rate of 3 radians/sec. What is the linear speed of a point on its rim, in meters per minute?
...