Mathematics, 25.12.2019 02:31 monica1217

Acone (vertex down) with height 10 inches and radius 2 inches is being filled with water at a constant rate of 2 in3/sec. how fast is the surface of the water rising when the depth of the water is 6 inches?

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Using implicit differentiation, it is found that the surface of the water is rising at a rate of 0.4775 inches per second.

The volume of a cone of radius r and height h is given by:

$V = \frac{\pi r^2h}{3}$

Applying implicit differentiation, the rate of change is given by:

$\frac{dV}{dt} = \frac{\pi}{3}\left[2r\frac{dr}{dt} + r^2\frac{dh}{dt}\right]$

The radius is constant, thus $\frac{dr}{dt} = 0$ .The remaining information given in the text is: $\frac{dV}{dt} = 2, r = 2$ .

Thus:

$\frac{dV}{dt} = \frac{\pi}{3}\left[2r\frac{dr}{dt} + r^2\frac{dh}{dt}\right]$

$2 = \frac{4\pi}{3}\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{6}{4\pi}$

$\frac{dh}{dt} = 0.4775$

The surface of the water is rising at a rate of 0.4775 inches per second.

A similar problem is given at link

Answer from: fefebat

Step-by-step explanation:

The volume for a cone is

$V=\frac{1}{3}\pi r^2h$

Let's look at the derivative of this to see what we need to fill into it against what we were given to fill into it. The derivative of the volume with respect to time is

$\frac{dV}{dt}=\frac{\pi }{3}[2r\frac{dr}{dt}h+r^2\frac{dh}{dt}]$

If this is the formula we are to use, we need a radius measure (we have one), a dr/dt (we don't have one), a height (we seem to have 2 of those!), and we are ultimately looking for dh/dt.

Because of the fact that we do not have a dr/dt value along with the fact that we have 2 values of height, we have to find a rewrite for that volume formula where the radius is in terms of the height. That way we don't need to worry about the rate of change of radius (becasue we weren't given one anyway!). The reason we were given that the height of the cone is 10 and the radius is 2 is so we can write the radius in terms of the height. Use proportions to do this:

$\frac{r}{h}=\frac{2}{10}$ Cross multiply to get 10r = 2h so

$r=\frac{1}{5}h$

We will now be able to express the volume in terms of height only:

$V=\frac{1}{3}\pi (\frac{1}{5}h)^2h$ which simplifies down to

$V=\frac{1}{3}\pi (\frac{1}{25})h^3$

This is what we will take the derivative of.

$\frac{dV}{dt}=\frac{1}{3}\pi (\frac{1}{25})3h^2\frac{dh}{dt}$

The 3's cancel, leaving a little bit of a simpler formula:

$\frac{dV}{dt}=\frac{1}{25}\pi h^2\frac{dh}{dt}$

We were given that dV/dt = 2 and we are to find dh/dt when h = 6. Filling in what we given then:

$2=\frac{\pi }{25}(6)^2\frac{dh}{dt}$ or

$2=\frac{36\pi }{25}\frac{dh}{dt}$

Multiply by the reciprocal of the fraction on the right and simplify to

$\frac{dh}{dt}=\frac{25}{18\pi }$ or, in decimal form .4420970641in/sec

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in fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.

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