Question

A tanker that ran aground is leaking oil that forms a circular slick about 0.2 foot thick. To estimate the rate​ dV/dt (in cubic feet per​ minute) at which oil is leaking from the​ tanker, it was found that the radius of the slick was increasing at 0.31 foot per minute ​(dR/dtequals0.31​) when the radius R was 400 feet. Find​ dV/dt, using pi almost equals 3.14 .

Answer 1

Answer:

$\frac{dV}{dt} = 155.82$ $\frac{ft^{3}}{min}$

Step-by-step explanation:

Since it is circular slick, it's volume can be modeled as,

$V = \pi R^{2}h$

Where R is the radius in feet and h is the thickness of slick.

taking derivative of the above equation with respect to time yields,

$\frac{dV}{dt} = \pi 2Rh \frac{dR}{dt}$

Where the rate of change of radius of the slick (dR/dt) is given,

$\frac{dV}{dt} = \pi 2(400)(0.2)(0.31)$

$\frac{dV}{dt} = 155.82$ $\frac{ft^{3}}{min}$

Therefore, the rate of change of volume is 155.82 cubic feet per minute.​

Answer 2

The value of dV/dt using pi almost equal to 3.14 gives; dV/dt = 155.82 ft³/min

dV/dt = 155.82 ft³/min

We are given;

Radius; R = 400 ft

Height; h = 0.2 ft

Rate of Increase of radius; dr/dt = 0.31 ft/min

Formula for Volume of a cylinder is;

V = πr²h

Differentiation of both sides of V and r with respect to t gives;

dV/dt = 2πrh(dr/dt)

Plugging in the relevant values gives;

dV/dt = 2π × 400 × 0.2 × 0.31

dV/dt = 155.82 ft³/min

Read more at; https://brainly.com/question/23611212