tutorial exercise\na cylindrical tank with radius 8 m is being filled with water at a rate of 2 m³/min. how…

tutorial exercise\na cylindrical tank with radius 8 m is being filled with water at a rate of 2 m³/min. how fast is the height of the water increasing?\nstep 1\nif h is the waters height, the volume of the water is v = πr²h. we must find dv/dt. differentiating both sides of the equation gives the following.\n\\(\\frac{dv}{dt}=\\frac{1}{16}\\times\\pi r^{2}\\frac{dh}{dt}\\)\nstep 2\nsubstituting for r, this becomes \\(\\frac{dv}{dt}=\\square\\times\\frac{dh}{dt}\\).\nsubmit\nskip (you cannot come back)

tutorial exercise\na cylindrical tank with radius 8 m is being filled with water at a rate of 2 m³/min. how fast is the height of the water increasing?\nstep 1\nif h is the waters height, the volume of the water is v = πr²h. we must find dv/dt. differentiating both sides of the equation gives the following.\n\\(\\frac{dv}{dt}=\\frac{1}{16}\\times\\pi r^{2}\\frac{dh}{dt}\\)\nstep 2\nsubstituting for r, this becomes \\(\\frac{dv}{dt}=\\square\\times\\frac{dh}{dt}\\).\nsubmit\nskip (you cannot come back)

Answer

Explanation:

Step1: Differentiate volume formula

Given $V = \pi r^{2}h$, using the product - rule (since $r$ is constant here, $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$).

Step2: Substitute the value of $r$

We know $r = 8$ m. Substituting $r = 8$ into $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$, we get $\frac{dV}{dt}=\pi\times8^{2}\frac{dh}{dt}=64\pi\frac{dh}{dt}$.

Step3: Solve for $\frac{dh}{dt}$

We are given that $\frac{dV}{dt}=2$ m³/min. So, $2 = 64\pi\frac{dh}{dt}$. Then $\frac{dh}{dt}=\frac{2}{64\pi}=\frac{1}{32\pi}$ m/min.

Answer:

$\frac{1}{32\pi}$ m/min