Respuesta :
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 [tex]\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}[/tex]
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 [tex]\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}[/tex]
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 [tex]0.0298\: \rm inch/sec\\[/tex] approximately.
How to calculate the instantaneous rate of growth of a function?
Suppose that a function is defined as;
[tex]y = f(x)[/tex]
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:
[tex]\dfrac{dy}{dx} = \dfrac{d(f(x))}{dx}[/tex]
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 = [tex]3/2 \: \rm inch^3/sec[/tex] = 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,
[tex]\dfrac{6}{6} = \dfrac{h}{r} \implies r = h[/tex], thus, radius of that cone of water will be of h inches too.
Thus, its volume is V = [tex]\dfrac{1}{3}\pi r^2h = \dfrac{1}{3} \pi h^3[/tex] cubic inches
It is given that:
[tex]\dfrac{dV}{dt} = 1.5 \: \rm inch^3/sec[/tex]
Taking the derivative, we get:
[tex]\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}[/tex]
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:
[tex]\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[/tex]
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:
[tex]\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\\\\[/tex]
Thus, the rate at which water level is rising when water is 4 inches deep is [tex]0.0298\: \rm inch/sec\\[/tex] approximately.
Learn more about instantaneous rate of change here:
https://brainly.com/question/13534870
