Respuesta :
Answer:
Altitude of the pile will increase by 16.56 cm per second.
Step-by-step explanation:
Sand is poured onto a surface at the rate = 13 cm³ per second
Or [tex]\frac{dV}{dt}=13[/tex]
It forms a conical pile with a diameter d cm and height of the pile = h cm
Here d = h
Volume of the pile [tex]V=\frac{1}{3}\times \pi r^{2}h[/tex]cm³per sec.
Since h = d = 2r [r is the radius of the circular base]
r = [tex]\frac{h}{2}[/tex]
[tex]V=\frac{1}{3}\pi (\frac{h}{2})^{2}h[/tex]
[tex]V=\frac{1}{3}\pi \frac{(h^{2})}{4}(h)[/tex]
[tex]V=\frac{1}{12}\pi h^{3}[/tex]
[tex]\frac{dV}{dt}=\frac{1}{12}\pi \times 3(h)^{2}\frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt}=\frac{1}{4}\pi \times h^{2}\times \frac{dh}{dt}[/tex]
Since [tex]\frac{dV}{dt}=13[/tex] cm³per sec.
13 = [tex]\frac{1}{4}\pi (1)^{2}\frac{dh}{dt}[/tex] [For h = 1 cm]
[tex]\frac{dh}{dt}=\frac{13\times4}{\pi }[/tex]
[tex]\frac{dh}{dt}=\frac{52}{3.14}[/tex]
[tex]\frac{dh}{dt}=16.56[/tex]cm per second.
Therefore, altitude of the pile will increase by 16.56 cm per second.
Measure of "how fast" is rate with respect to time passed. At 1 cm height of pile, the altitude of the pile is rising at the instantaneous speed 16.552 cubic cm/sec.
How to calculate the instantaneous speed 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:
[tex]\dfrac{dy}{dx} = \dfrac{d(f(x))}{dx}[/tex]
For the given case, we want the height increment rate with respect to time passed. For that, we will have to first find the height achieved by the pile in terms of the time passed.
Since it is given that:
The sand is being poured onto a surface at rate of [tex]13 cm^3 / sec[/tex] speed.
That means, as each second passes, 13 cubic centimetres of sand is added to that conical shape.
Also, it is given that:
Diameter of base of the cone being formed = Altitude of the cone
Thus, radius of the base of the cone = half of altitude of the cone
or
r = h/2
Volume of such cone would be
[tex]V = \dfrac{1}{3} \pi r^2 \times h = \dfrac{1}{3} \pi \dfrac{h^3}{4} = \dfrac{\pi h^3}{12} \: \rm unit^3[/tex]
It is known that:
[tex]\dfrac{dV}{dt} = 13[/tex] (cubic cm/sec) (rate is constant, thus, same on instantaneous scale too)
Thus, we get:
[tex]\dfrac{dV}{dt} = \dfrac{d(\pi h^3/12)}{dt} = \dfrac{\pi}{12} \times \dfrac{dh^3}{dt} = \dfrac{\pi}{12} \times 3h^2\dfrac{dh}{dt}\\\\13 = \dfrac{\pi h^2}{4} \dfrac{dh}{dt}\\\\\dfrac{dh}{dt} = \dfrac{52}{\pi \times h^2}[/tex]
This equation above is the instantaneous rate of change of height with respect to time. It is in unit as cubic cm/sec
When height = 1 cm, we get h = 1 or
[tex]\dfrac{dh}{dt} = \dfrac{52}{\pi \times h^2}\\\\\dfrac{dh}{dt}|_{h=1} = \dfrac{52}{\pi \times 1} =\dfrac{52}{\pi} \approx 16.552 \: \rm cm^3/sec[/tex]
Thus, at 1 cm height of pile, the altitude of the pile is rising at the instantaneous speed 16.552 cubic cm/sec.
Learn more about instantaneous rate here:
https://brainly.com/question/4746888
