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3 years ago

Please help me with this problem. I'm stuck!!

Mathematics

1 answer:

mihalych1998 [28]3 years ago

6 0

$\bf S(t)=4\pi \left( \cfrac{3}{8}t^4 \right)^2\impliedby \textit{distributing the exponent}
\\\\\\
S(t)=4\pi \left( \cfrac{3^2}{8^2}t^{4\cdot 2} \right)\implies S(t)=4\pi \cdot \cfrac{9t^8}{64}\implies S(t)=\cfrac{4}{64}\cdot \cfrac{9\pi t^8}{1}
\\\\\\
S(t)=\cfrac{1}{16}\cdot \cfrac{9\pi t^8}{1}\implies S(t)=\cfrac{9\pi t^8}{16}$

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2 years ago

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nirvana33 [79]

Answer:

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And replacing we got:

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Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

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Answer:

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