1. Problem (a): A balloon is rising vertically at 3 m/s. A boy cycles beneath it at 5 m/s. When the balloon is 40 m high, find the rate of change of the distance between them. 2. Let $x$ be the horizontal distance of the boy from the point beneath the balloon, and $y$ the height of the balloon. 3. Given: $\frac{dy}{dt} = 3$ m/s, $\frac{dx}{dt} = 5$ m/s, and at the instant, $y=40$ m. 4. The distance between boy and balloon is $z = \sqrt{x^2 + y^2}$. 5. Differentiate with respect to time $t$: $$\frac{dz}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}$$ 6. At $t=0$, the boy is directly beneath the balloon, so $x=0$. After some time $t$, $x = 5t$ and $y = 40 + 3t$. 7. To find $\frac{dz}{dt}$ at the instant when $y=40$, we consider $x=0$ (since the boy is just passing beneath). 8. Substitute $x=0$, $y=40$, $\frac{dx}{dt}=5$, $\frac{dy}{dt}=3$: $$\frac{dz}{dt} = \frac{0 \cdot 5 + 40 \cdot 3}{\sqrt{0^2 + 40^2}} = \frac{120}{40} = 3$$ 9. So, the distance between the boy and balloon is increasing at 3 m/s at that instant. --- 10. Problem (b): Water leaks from an inverted conical tank at $\frac{dV}{dt} = -2$ m³/min. Tank height $H=12$ m, radius $R=6$ m. Find rate of decrease of water depth $h$ when $h=4$ m. 11. Volume of cone: $V = \frac{1}{3} \pi r^2 h$. 12. Relate $r$ and $h$ by similar triangles: $\frac{r}{h} = \frac{R}{H} = \frac{6}{12} = \frac{1}{2} \Rightarrow r = \frac{h}{2}$. 13. Substitute $r$ into $V$: $$V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{4} h = \frac{\pi}{12} h^3$$ 14. Differentiate volume w.r.t. time: $$\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$ 15. Given $\frac{dV}{dt} = -2$ m³/min and $h=4$ m, solve for $\frac{dh}{dt}$: $$-2 = \frac{\pi}{4} (4)^2 \frac{dh}{dt} = \frac{\pi}{4} \cdot 16 \frac{dh}{dt} = 4\pi \frac{dh}{dt}$$ 16. Thus: $$\frac{dh}{dt} = \frac{-2}{4\pi} = -\frac{1}{2\pi}$$ 17. The depth of water is decreasing at a rate of $\frac{1}{2\pi}$ m/min when $h=4$ m. Final answers: (a) Rate of change of distance between boy and balloon is $3$ m/s. (b) Rate of decrease of water depth is $\frac{1}{2\pi}$ m/min.