Answer :
Answer: The height of the oil changing at the rate of [tex]\dfrac{200}{49\pi}\ inches/sec[/tex]
Step-by-step explanation:
Since we have given that
volume is increasing at a rate of 150 cubic inches per second.
Height of cylinder = 10 times of radius.
As we know the formula for "Volume":
[tex]V=\pi r^2h\\\\V=\pi (\dfrac{h}{10})^2\times 10r\\\\V=\dfrac{1}{100}\pi h^3[/tex]
We will derivative it w.r.t 't'.
So, it becomes,
[tex]\dfrac{dv}{dt}=\dfrac{3h^2\pi}{100}\times \dfrac{dh}{dt}\\\\150=\dfrac{3h^2\pi}{100}\dfrac{dh}{dt}\\\\\dfrac{150\times 100}{3\times 35\times 35\times \pi}=\dfrac{dh}{dt}\\\\\dfrac{200}{49\pi}=\dfrac{dh}{dt}[/tex]
Hence, the height of the oil changing at the rate of [tex]\dfrac{200}{49\pi}\ inches/sec[/tex]
Answer:
[tex]\dot h = 1.299\,\frac{in}{s}[/tex]
Step-by-step explanation:
The volume of the cylindrical container is:
[tex]V = \pi\cdot r^{2}\cdot h[/tex]
Where:
[tex]r[/tex] - Radius, in inches.
[tex]h[/tex] - Height, in inches.
The rate of change of the volume is:
[tex]\dot V = 2\pi\cdot r \cdot h \cdot \dot r + \pi\cdot r^{2}\cdot \dot h[/tex]
[tex]\dot V = \pi \cdot r \cdot (2\cdot h \cdot \dot r + r \cdot \dot h)[/tex]
[tex]\dot V = 0.1\pi \cdot h \cdot (0.2 \cdot h \cdot \dot h + 0.1\cdot h \cdot \dot h)[/tex]
[tex]\dot V = 0.03\pi\cdot h^{2} \cdot \dot h[/tex]
The rate of change of the height is:
[tex]\dot h = \frac{\dot V}{0.03\pi\cdot h^{2}}[/tex]
[tex]\dot h = \frac{150\,\frac{in^{3}}{s} }{0.03\pi\cdot (35\,in)^{2}}[/tex]
[tex]\dot h = 1.299\,\frac{in}{s}[/tex]