A cylindrical tank with radius 5 cm is being filled with water at rate of 3 cm^3 per min. how fast is the height of the water increasing?
I dont want this question solved, but please help me correct my working out:
- radius = 5
- dv/dt = 3
- dh/dt = dh/dv * dv/dt
- v=(pi)(r^2)(h)
because r is constant you could write: V=(pi)(5^2)(h) and then find the derivative…
but is there an alternate method where we can derive dv/dh without first substituting r=5??
Best Answer
There is no other way to find $\frac{dv}{dh}$ without substituting $r = 5$.
To solve the question:
$$V = \pi r^2h$$
Since r is constant: $$V = \pi 25h$$ $$\frac{dV}{dt} = 25 \pi*\frac{dh}{dt}$$ $$3 = 25 \pi*\frac{dh}{dt}$$ $$\frac{dh}{dt} = \frac{3}{25\pi} \text{cm/min}$$