Respuesta :
Answer:
Step-by-step explanation:
This is related rates in calculus, as far as I can tell from the info you've provided. It's nice to see students are still trying to conquer its vagueness!
Start with the formula for volume of a cylinder:
[tex]V=\pi r^2h[/tex] and then find its derivative with respect to time:
[tex]\frac{dV}{dt}=\pi (r^2\frac{dh}{dt}+2r\frac{dr}{dt}h)[/tex]
From that formula it looks like we need a value for dV/dt, r, dh/dt, and h; dr/dt is our unknown. We have that
dV/dt = -171
dh/dt = 6
h = 6 and
V = 1254
We need a value for r. We can find it using the last 2 values listed above in the volume formula:
[tex]1254=\pi r^2(6)[/tex] and
[tex]\frac{1254}{6\pi } =r^2[/tex] and
[tex]66.52676621=r^2[/tex] and
r = 8.156394
NOW we have everything we need to fill in the derivative for the volume:
[tex]-171=66.52676621(6)\pi +6(2)(8.156394)\pi \frac{dr}{dt}[/tex] which simplifies to
[tex]-171=1254+307.4888096\frac{dr}{dt}[/tex] and
[tex]-1425=307.488096\frac{dr}{dt}[/tex] so
[tex]\frac{dr}{dt}=-4.634[/tex] ft/sec
The rate of change of the radius with time is a partial derivative of the rate
of change of volume with time.
At the instant the height of the cylinder is 6 feet is the radius is decreasing
at a rate of approximately 4.634 ft./s.
Reasons:
The given parameters are;
Rate of increase of the height of the cylinder, [tex]\dfrac{dh}{dt}[/tex] = 6 ft./s
Rate of decrease of the volume of the cylinder, [tex]\dfrac{dV}{dt}[/tex] = 171 ft.³/s
Height of the cylinder, h = 6 ft.
Volume of the cylinder, V = 1254 ft.³
The formula for the volume of the cylinder, V = π·r²·h
Solution:
By partial differentiation, we have;
[tex]\dfrac{dz}{du} = \mathbf{\dfrac{\partial z}{\partial x} \cdot \dfrac{dx}{du} + \dfrac{\partial z}{\partial y} \cdot \dfrac{dy}{du}}[/tex]
Therefore;
[tex]\dfrac{dV}{dt} = \dfrac{\partial (\pi \cdot r^2 \cdot h) }{\partial r} \cdot \dfrac{dr}{dt} + \dfrac{\partial (\pi \cdot r^2 \cdot h)}{\partial h} \cdot \dfrac{dh}{dt} = \pi \cdot \left(2 \cdot r \cdot h \cdot \dfrac{dr}{dt} + r^2 \cdot \dfrac{dh}{dt} \right)[/tex]
[tex]\dfrac{dV}{dt} = \mathbf{\pi \cdot \left(2 \cdot r \cdot h \cdot \dfrac{dr}{dt} + r^2 \cdot \dfrac{dh}{dt} \right)}[/tex]
The value of the radius, r, in the given condition is therefore;
[tex]r = \sqrt{\dfrac{V}{\pi \cdot h} }[/tex]
Which gives;
[tex]r = \sqrt{\dfrac{1254}{\pi \times 6} } = \dfrac{\sqrt{209 \cdot \pi} }{\pi}[/tex]
Therefore, we get;
[tex]-171 = \mathbf{\pi \cdot \left(2 \times \dfrac{\sqrt{209 \cdot \pi} }{\pi} \times 6 \times \dfrac{dr}{dt} + \left(\dfrac{\sqrt{209 \cdot \pi} }{\pi} \right)^2 \times6 \right)}[/tex]
[tex]\dfrac{dr}{dt} =\dfrac{-171-1254}{12 \cdot \sqrt{209 \cdot \pi} } = \dfrac{-475}{4 \cdot \sqrt{209 \cdot \pi} } \approx -4.634[/tex]
[tex]\mathrm {The \ rate \ of \ change \ of \ the \ radius \ with \ time} \ \dfrac{dr}{dt} \approx -4.634 \ ft./s[/tex]
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