Respuesta :
Answer:
The rate at which the height of the pile is [tex]\dfrac{2}{5\pi}[/tex].
Explanation:
Given that,
Height = radius = 5 m
Rate = 10 m³/s
We need to calculate the rate at which the height of the pile
Using formula of volume
[tex]V=\dfrac{1}{3}\pi\times r^2\times h[/tex]
r = h,
On differentiating
[tex]\dfrac{dV}{dt}=\pi\times h^2\times\dfrac{dh}{dt}[/tex]
Put the value into the formula
[tex]10=\pi\times25\times\dfrac{dh}{dt}[/tex]
[tex]\dfrac{dh}{dt}=\dfrac{10}{25\pi}[/tex]
[tex]\dfrac{dh}{dt}=\dfrac{2}{5\pi}[/tex]
Hence, The rate at which the height of the pile is [tex]\dfrac{2}{5\pi}[/tex].
Answer:
0.127 m/s
Explanation:
Let the diameter of the conical pile is d.
height of the conical pile = d/2
dV/dt = 10m^3/s
h = 5 m
then, d = 2 x h = 2 x 5 = 10 m
Volume of cone is given by
[tex]V=\frac{1}{3}\pi r^{2}h[/tex]
where, r is the radius of the pile.
r = d/2 and h = d/2 , it means r = h
[tex]V=\frac{1}{3}\pi \times h^{3}[/tex]
Differentiate with respect to t on both the sides
[tex]\frac{dV}{dh}=\frac{1}{3}\pi \times 3h^{2}\times \frac{dh}{dt}[/tex]
by substituting the values
[tex]10=\frac{1}{3}\pi \times 3\times 5\times 5\times \frac{dh}{dt}[/tex]
dh/dt = 0.127 m/s
Thus, the rate of change of height is 0.127 m/s.
