Question

Water is dropped at the rate of 2 m³/sec into a cone of semi-vertical angle 45°. The rate at which circumference of water top surface changes when height of the water in the cone is 2 meter, is __________.

Answer 1

1

Answer 2

Core Solution:

1. Relate radius and height using the semi-vertical angle: Let $r$ be the radius of the water surface and $h$ be the height of the water in the cone. The semi-vertical angle $\alpha = 45^\circ$. From the geometry of the cone, $\tan \alpha = \frac{r}{h}$. Given $\alpha = 45^\circ$, we have $\tan 45^\circ = 1$. Therefore, $1 = \frac{r}{h} \implies r = h$.

2. Express the volume of water in terms of height: The volume of water in the cone is given by $V = \frac{1}{3}\pi r^2 h$. Substitute $r=h$ into the volume formula: $V = \frac{1}{3}\pi (h)^2 h = \frac{1}{3}\pi h^3$.

3. Differentiate the volume with respect to time: We are given the rate at which water is dropped into the cone, $\frac{dV}{dt} = 2$ m³/sec. Differentiate $V = \frac{1}{3}\pi h^3$ with respect to time $t$: $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{3}\pi h^3\right) = \frac{1}{3}\pi \cdot 3h^2 \frac{dh}{dt} = \pi h^2 \frac{dh}{dt}$.

4. Calculate the rate of change of height ($\frac{dh}{dt}$): Substitute the given values $\frac{dV}{dt} = 2$ m³/sec and $h = 2$ m into the differentiated volume equation: $2 = \pi (2)^2 \frac{dh}{dt}$ $2 = 4\pi \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi}$ m/sec.

5. Express the circumference of the water top surface in terms of height: The circumference of the water top surface is $C = 2\pi r$. Since $r=h$, we have $C = 2\pi h$.

6. Differentiate the circumference with respect to time: To find the rate at which the circumference changes, differentiate $C = 2\pi h$ with respect to time $t$: $\frac{dC}{dt} = \frac{d}{dt}(2\pi h) = 2\pi \frac{dh}{dt}$.

7. Calculate the rate of change of circumference ($\frac{dC}{dt}$): Substitute the value of $\frac{dh}{dt}$ found in step 4: $\frac{dC}{dt} = 2\pi \left(\frac{1}{2\pi}\right)$ $\frac{dC}{dt} = 1$ m/sec.

The rate at which the circumference of the water top surface changes is 1 m/sec.