the volume of a sphere is increasing at a constant rate of 91 cubic inches per minute. at the instant when…

the volume of a sphere is increasing at a constant rate of 91 cubic inches per minute. at the instant when the volume of the sphere is 37 cubic inches, what is the rate of change of the surface area of the sphere? the volume of a sphere can be found with the equation $v = \\frac{4}{3}\\pi r^3$ and the surface area can be found with $s = 4\\pi r^2$. round your answer to three decimal places (if necessary).

the volume of a sphere is increasing at a constant rate of 91 cubic inches per minute. at the instant when the volume of the sphere is 37 cubic inches, what is the rate of change of the surface area of the sphere? the volume of a sphere can be found with the equation $v = \\frac{4}{3}\\pi r^3$ and the surface area can be found with $s = 4\\pi r^2$. round your answer to three decimal places (if necessary).

Answer

Explanation:

Step1: Relate ( V ) and ( r ) at ( V = 37 )

We know ( V=\frac{4}{3}\pi r^{3} ). When ( V = 37 ), we solve for ( r ): [ 37=\frac{4}{3}\pi r^{3} ] First, multiply both sides by ( \frac{3}{4\pi} ): [ r^{3}=\frac{3\times37}{4\pi}=\frac{111}{4\pi} ] Then take the cube - root: [ r=\sqrt[3]{\frac{111}{4\pi}}\approx\sqrt[3]{\frac{111}{12.5664}}\approx\sqrt[3]{8.834}\approx2.068 ]

Step2: Differentiate ( V ) and ( S ) with respect to time ( t )

Differentiate ( V = \frac{4}{3}\pi r^{3} ) with respect to ( t ) using the chain rule (( \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt} )) Differentiate ( S = 4\pi r^{2} ) with respect to ( t ) using the chain rule (( \frac{dS}{dt}=8\pi r\frac{dr}{dt} ))

Step3: Solve for ( \frac{dr}{dt} ) from the volume equation

We know that ( \frac{dV}{dt} = 91 ) cubic inches per minute. From ( \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt} ), we can solve for ( \frac{dr}{dt} ): [ \frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^{2}} ] Substitute ( \frac{dV}{dt}=91 ) and ( r\approx2.068 ): [ \frac{dr}{dt}=\frac{91}{4\pi\times(2.068)^{2}}\approx\frac{91}{4\pi\times4.276}\approx\frac{91}{53.76}\approx1.693 ]

Step4: Substitute ( \frac{dr}{dt} ) into the surface - area derivative

Now, substitute ( r\approx2.068 ) and ( \frac{dr}{dt}\approx1.693 ) into ( \frac{dS}{dt}=8\pi r\frac{dr}{dt} ): [ \frac{dS}{dt}=8\pi\times2.068\times1.693 ] [ \frac{dS}{dt}\approx8\times3.1416\times2.068\times1.693 ] [ \frac{dS}{dt}\approx25.1328\times2.068\times1.693 ] [ \frac{dS}{dt}\approx25.1328\times3.503 ] [ \frac{dS}{dt}\approx88.04 ] (We can also do it in a more combined way. From ( \frac{dS}{dt}=8\pi r\frac{dr}{dt} ) and ( \frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^{2}} ), we can substitute ( \frac{dr}{dt} ) into ( \frac{dS}{dt} ): ( \frac{dS}{dt}=8\pi r\times\frac{\frac{dV}{dt}}{4\pi r^{2}}=\frac{2\frac{dV}{dt}}{r} ) Substitute ( \frac{dV}{dt} = 91 ) and ( r\approx2.068 ): ( \frac{dS}{dt}=\frac{2\times91}{2.068}=\frac{182}{2.068}\approx88.01 ) (The slight difference is due to rounding of ( r ) at different steps))

Answer:

The rate of change of the surface area is approximately (\boldsymbol{88.01}) square inches per minute.