Water is poured into a conical paper cup at the rate of 3/2 in3/sec. If the cup is 6 inches tall and the top has a radius of 6 inches, how fast is the water level rising when the water is 4 inches deep?

so.. hmm notice the picture below

r = h, due to similar triangles ratio, whatever "h" may be

now, we're given how much water is being poured into the cup per sec
namely dv/dt the volume's rate

so.. hmmm \(\bf \textit{volume of a cone}=V=\cfrac{\pi r^2 h}{3}\quad however\implies r=h \\\\\\ thus\implies V=\cfrac{\pi h^2 h}{3}\implies V=\cfrac{\pi h^3}{3}\\\\ ------------------------\\\\ now \\\\\\ \cfrac{dv}{dt}=\cfrac{\pi }{3}\cdot 3h^2\cdot \cfrac{dh}{dt}\implies \cfrac{dv}{dt}=\pi h^2\cdot \cfrac{dh}{dt} \\\\\\ \cfrac{\frac{dv}{dt}}{\pi h^2}=\cfrac{dh}{dt}\quad \begin{cases} \frac{dv}{dt}=\frac{3}{2}\\\\ \left. \frac{dh}{dt} \right|_{h=4} \end{cases}\implies \cfrac{\frac{3}{2}}{\pi 4^2}=\cfrac{dh}{dt}\)

astha8579 astha8579 · Answer 2 · 2022-02-26T01:55:51-05:00

Rate of something is always with compared to other quantity. The rate at which water level is rising when water is 4 inches deep is \(0.0298\: \rm inch/sec\\\) approximately.

How to calculate the instantaneous rate of growth of a function?

Suppose that a function is defined as;

\(y = f(x)\)

Then, suppose that we want to know the instantaneous rate of the growth of the function with respect to the change in x, then its instantaneous rate is given as:

\(\dfrac{dy}{dx} = \dfrac{d(f(x))}{dx}\)

For the given case, its given that:

The height of conical paper cup = 6 inches
Radius of top = 6 inches.
The rate at which water is being poured = \(3/2 \: \rm inch^3/sec\) = 1.5 cubic inch/sec

Suppose that the water level is at h units, then the volume of the water contained at that level is given by the volume of cone which has height h inches and the radius= radius of the circular water film on the top.

Since the radius to height ratio will stay common due to same slope of both cones, thus,

\(\dfrac{6}{6} = \dfrac{h}{r} \implies r = h\), thus, radius of that cone of water will be of h inches too.

Thus, its volume is V = \(\dfrac{1}{3}\pi r^2h = \dfrac{1}{3} \pi h^3\) cubic inches

It is given that:

\(\dfrac{dV}{dt} = 1.5 \: \rm inch^3/sec\)

Taking the derivative, we get:

\(\dfrac{dV}{dt} = \dfrac{\dfrac{1}{3} \pi h^3}{dt} = \dfrac{1}{3} \pi 3h^2 \dfrac{dh}{dt} = \pi h^2\dfrac{dh}{dt}\)

where dh/dt is the rate at which the height of the cone made by water is increasing(which is what we need). Using the rate of volume increment and the above equation, we get:

\(\dfrac{dV}{dt} = \pi h^2\dfrac{dh}{dt}\\\\1.5 = \dfrac{\pi h^2}\dfrac{dh}{dt} \\\\\dfrac{dh}{dt} = \dfrac{1.5}{\pi h^2}\: \rm inch/sec\)

This is the rate of increment of height of the water in the given cone when height is h inches.

For h = 4, we get:

\(\dfrac{dh}{dt} = \dfrac{1.5}{\pi h^2}\: \rm inch^3/sec\\\\\dfrac{dh}{dt}|_{h=4} = \dfrac{1.5}{\pi 4^2} \approx 0.0298\: \rm inch/sec\\\\\)

Thus, the rate at which water level is rising when water is 4 inches deep is \(0.0298\: \rm inch/sec\\\) approximately.

Learn more about instantaneous rate of change here:

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