Limits and Derivatives: Real-World Scenarios (2)

Question

A water tank in the shape of an inverted cone (vertex down) has height $10$ m and top radius $5$ m. Water drains out at a rate of $0.5\,\text{m}^3$/min. How fast is the water level falling when the depth is $4$ m?

A classic related-rates problem on the Class 11 / JEE Main syllabus.

Solution — Step by Step

Final answer: water level falls at $1/(8\pi) \approx 0.04$ m/min.

Why This Works

Related rates work because all variables in the geometric relation are functions of time. Differentiating both sides of $V(t) = f(h(t))$ uses the chain rule to link $dV/dt$ with $dh/dt$ via the geometry.

The similar-triangles step is what makes the cone tractable — by expressing $r$ in terms of $h$, we eliminate one variable before differentiating. Otherwise, we’d have a partial-derivatives problem.

Alternative Method

Implicit differentiation. Keep $V = \tfrac{1}{3}\pi r^2 h$ with the constraint $r = h/2$. Differentiate both: $dV/dt = \tfrac{1}{3}\pi(2r\cdot dr/dt \cdot h + r^2\cdot dh/dt)$ and $dr/dt = (1/2)dh/dt$. Substitute: get the same expression for $dh/dt$. Slower but generalises better.

Common Mistake

Substituting $h = 4$ into $V = \pi h^3/12$ before differentiating. That kills the variable and you get $dh/dt = 0$. Always differentiate first, then substitute the specific value.