The volume of a cylinder is increasing at a rate of 10π cubic meters per hour.
The height of the cylinder is fixed at 5 meters.
At a certain instant, the volume is 80π cubic meters.
What is the rate of change of the surface area of the cylinder at that instant (in square meters per hour)?

Respuesta :

Answer:

13π/2 m²/h

Step-by-step explanation:

Volume of a cylinder is:

V = πr²h

If h is a constant, then taking derivative of V with respect to time:

dV/dt = 2πrh dr/dt

Surface area of a cylinder is:

A = 2πr² + 2πrh

Taking derivative with respect to time:

dA/dt = (4πr + 2πh) dr/dt

Given that dV/dt = 10π, V = 80π, and h = 5, we need to find dA/dt.  But first, we need to find r and dr/dt.

V = πr²h

80π = πr² (5)

r = 4

dV/dt = 2πrh dr/dt

10π = 2π (4) (5) dr/dt

dr/dt = 1/4

dA/dt = (4πr + 2πh) dr/dt

dA/dt = (4π (4) + 2π (5)) (1/4)

dA/dt = 13π/2

The surface area of the cylinder is increasing at 13π/2 m²/h.

The required surface area of the cylinder is [tex]\frac{13}{2}[/tex] m²/h.

Given that,

The volume of the cylinder increase with the rate of 10π cubic meters per hour.

The height of the cylinder is fixed at 5 meters.

At a certain instant, the volume is 80π cubic meters.

We have to determine,

What is the rate of change of the surface area of the cylinder at that instant.

According to the question,

Height of cylinder = 5m

Volume of cylinder increase with rate =  10π cubic meters per hour.

At certain instant volume become =  8cubic meters per hour.

Volume of cylinder is given as,

V = πr²h

Where h is a constant,

[tex]v = \pi r^{2} h\\\\80\pi = \pi r^{2}. (5)\\\\\frac{80\pi }{5\pi } = r^{2} \\\\r^{2} = 16\\\\r = 4[/tex]

Then,  taking derivative of V with respect to time:

[tex]\frac{dv}{dt} = 2\pi rh.\frac{dr}{dt} \\\\[/tex]

Where, dv\dt = 10π, V = 80π, and h = 5,

Then,

[tex]10\pi = 2\pi (4)(5). \frac{dr}{dt} \\\\10\pi = 40\pi \frac{dr}{dt} \\\\\frac{10\pi }{40\pi } = \frac{dr}{dt} \\\\\\\frac{dr}{dt} = \frac{1}{4}[/tex]

Then,

Surface area of a cylinder is define as,

[tex]A = 2\pi r^{2} + 2\pi rh\\\\\ \frac{da}{dt} = (4\pi r + 2\pi h)\frac{dr}{dt} \\\\\frac{da}{dt} = (4\pi (4) + 2\pi (5))\times\frac{1}{4} \\\\\frac{da}{dt} = \frac{26\pi }{4} \\\\\frac{da}{dt} = \frac{13\pi }{2}[/tex]

Hence, The required surface area of the cylinder is [tex]\frac{13}{2}[/tex] m²/h.

For more information about Surface area click the link given below.

https://brainly.com/question/16177070