Q.(a) A conical water tank with vertex down of 12 meters height has a radius of 5 meters at the top. If water flows into the tank at a rate 10 cubic m/min, how fast is the depth of the water increasing when the water is 8 metres deep ? OR

(a) Uses similar triangles to relate the cone's radius and height, then related rates on the volume formula; (b) uses the cross product (rather than dot product) of two unit vectors to derive the sine-difference formula. Both alternatives answered below.

(a) Related rates in a conical tank

1. Similar triangles. The cone has height $12$ m and top radius $5$ m; at water depth $h$, the water surface radius $r$ satisfies $\dfrac rh=\dfrac5{12}\Rightarrow r=\dfrac{5h}{12}$.

2. Volume in terms of $h$ alone.

$V=\dfrac13\pi r^2h=\dfrac13\pi\left(\dfrac{5h}{12}\right)^2h=\dfrac13\pi\cdot\dfrac{25h^2}{144}\cdot h=\dfrac{25\pi h^3}{432}$

3. Differentiate with respect to time.

$\dfrac{dV}{dt}=\dfrac{25\pi}{432}\cdot3h^2\dfrac{dh}{dt}=\dfrac{75\pi h^2}{432}\dfrac{dh}{dt}=\dfrac{25\pi h^2}{144}\dfrac{dh}{dt}$

4. Substitute $\dfrac{dV}{dt}=10$ m³/min, $h=8$ m:

$10=\dfrac{25\pi(64)}{144}\dfrac{dh}{dt}=\dfrac{1600\pi}{144}\dfrac{dh}{dt}=\dfrac{100\pi}9\dfrac{dh}{dt}$

5. Solve.

$\dfrac{dh}{dt}=\dfrac{10\times9}{100\pi}=\dfrac{90}{100\pi}=\dfrac{9}{10\pi}\ \text{m/min}\approx0.29\ \text{m/min}$

(b) Vector proof of $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

1. Unit vectors. Let $\hat a=\cos\alpha\,\hat i+\sin\alpha\,\hat j$ and $\hat b=\cos\beta\,\hat i+\sin\beta\,\hat j$ (angles $\alpha,\beta$ from the $x$-axis).

2. Cross product $\hat b\times\hat a$ by components (both vectors lie in the $xy$-plane, so the result is along $\hat k$):

$\hat b\times\hat a=(\cos\beta\,\hat i+\sin\beta\,\hat j)\times(\cos\alpha\,\hat i+\sin\alpha\,\hat j)=(\cos\beta\sin\alpha-\sin\beta\cos\alpha)\hat k=(\sin\alpha\cos\beta-\cos\alpha\sin\beta)\hat k$ …