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Chapter 6 · Application of Derivatives

Miscellaneous · Q16 (MCQ): Wheat into a tank of radius 10 m at 314 m³/h: depth rises at about 1 m/h

Miscellaneous ExerciseRate of change of quantities · Section 6.2 · Miscellaneous Exercise
The question

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of

(A) 1 m/h    (B) 0.1 m/h    (C) 1.1 m/h    (D) 0.5 m/h

Exam answer

Let the volume of wheat be VV m³ and its depth hh m at time tt hours. Then

V=π(10)2h=100πhV = \pi(10)^2 h = 100\pi h

Differentiating w.r.t. tt⁠, dVdt=100πdhdt\frac{dV}{dt} = 100\pi\frac{dh}{dt}⁠. With dVdt=314\frac{dV}{dt} = 314⁠,

dhdt=314100π=3.14π≈1\frac{dh}{dt} = \frac{314}{100\pi} = \frac{3.14}{\pi} \approx 1

Answer: (A) 1 m/h

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

In a cylinder the cross-section is the same at every level, so the volume of wheat is (area of the base) ×\times (depth): V=100πhV = 100\pi h⁠. Differentiating with respect to time links the two rates: dVdt=100π dhdt\frac{dV}{dt} = 100\pi\,\frac{dh}{dt}⁠. The book's number 314 is 100π100\pi with π≈3.14\pi \approx 3.14⁠, so the depth rises at about 1 m/h.

Given, to find, and the plan
Given
A vertical cylindrical tank of radius r=10r = 10 m. Wheat enters at dVdt=314\frac{dV}{dt} = 314 m³/h, where VV is the volume of wheat (m³) at time tt (hours).
To find
dhdt\frac{dh}{dt}⁠, the rate at which the depth hh (m) of the wheat increases, and the matching option.
Plan
(1) Write VV in terms of hh⁠. (2) Differentiate with respect to tt (chain rule). (3) Substitute and solve for dhdt\frac{dh}{dt}⁠. (4) Examine every option.

Basics you need

1Volume of a cylinder

V=πr2hV = \pi r^2 h⁠: base area πr2\pi r^2 times height (here, the depth of wheat). Basics: area and volume formulas

2Related rates

If VV depends on hh and hh depends on tt⁠, then dVdt=dVdh⋅dhdt\frac{dV}{dt} = \frac{dV}{dh}\cdot\frac{dh}{dt} (chain rule). Basics: related rates

3Rate of change as a derivative (Section 6.2)

"The rate at which hh increases" means dhdt\frac{dh}{dt}⁠; a positive value means hh is increasing. Basics: rate of change

Solution

  1. Volume of wheat in terms of its depth

    The wheat fills the tank up to depth hh⁠, forming a cylinder of radius 10 m and height hh⁠:

    V=π(10)2h=100πh.V = \pi (10)^2 h = 100\pi h.
    10 m h wheat V = π r² h with r = 10 m fixed so V = 100π h dV/dt = 314 m³/h dh/dt = 314/(100π) ≈ 1 m/h
    The radius stays 10 m at every level, so each extra metre of depth needs the same 100π≈314100\pi \approx 314 m³ of wheat.
    The radius is fixed by the tank; only the depth changes as wheat pours in. So VV is a function of hh alone, and the constant 100π100\pi is the area of the base in m².
    V=πr2hV = \pi r^2 h with r=10r = 10⁠. Basics: area and volume formulas

  2. Differentiate with respect to time

    Both VV and hh change with time tt⁠. Differentiating V=100πhV = 100\pi h with respect to tt (chain rule):

    dVdt=100π dhdt.\frac{dV}{dt} = 100\pi\,\frac{dh}{dt}.
    dVdh=100π\frac{dV}{dh} = 100\pi is a constant, so the chain rule simply multiplies dhdt\frac{dh}{dt} by it. We differentiate first and substitute numbers afterwards; here the relation is the same at every depth.
    Chain rule: dVdt=dVdh⋅dhdt\frac{dV}{dt} = \frac{dV}{dh}\cdot\frac{dh}{dt}⁠. Basics: related rates

  3. Substitute dVdt=314\frac{dV}{dt} = 314 and solve

    314=100π dhdtdhdt=314100π=3.14π m/h.\begin{aligned} 314 &= 100\pi\,\frac{dh}{dt} \\ \frac{dh}{dt} &= \frac{314}{100\pi} = \frac{3.14}{\pi}\ \text{m/h}. \end{aligned}

    With the usual approximation π≈3.14\pi \approx 3.14⁠, this is 3.143.14=1\frac{3.14}{3.14} = 1 m/h. (With the exact π\pi⁠, 3.14π≈0.9995\frac{3.14}{\pi} \approx 0.9995 m/h, which still rounds to 1 m/h.) The rate is positive, so the depth is increasing, at about 1 m/h.

    The figure 314 was chosen as 100×3.14100 \times 3.14⁠, so that π\pi cancels when it is taken as 3.14. Units check: m³/h divided by m² gives m/h.
    Dividing both sides by 100π≠0100\pi \ne 0⁠. Basics: rate of change

  4. Examine every option

    Since dVdt=100π dhdt≈314 dhdt\frac{dV}{dt} = 100\pi\,\frac{dh}{dt} \approx 314\,\frac{dh}{dt}⁠, each option fixes the filling rate it would need:

    • (A) 1 m/h: needs 314×1=314314 \times 1 = 314 m³/h. This is the given rate, so (A) is correct.
    • (B) 0.1 m/h: needs 314×0.1=31.4314 \times 0.1 = 31.4 m³/h, ten times too little. It comes from a factor-of-10 slip, such as dividing by 1000π1000\pi instead of 100π100\pi⁠.
    • (C) 1.1 m/h: needs 314×1.1=345.4314 \times 1.1 = 345.4 m³/h, more than is being poured in. No correct reading of the data gives it.
    • (D) 0.5 m/h: needs 314×0.5=157314 \times 0.5 = 157 m³/h, half the given rate. It results from doubling the area, for example dividing by 2πr2=200π2\pi r^2 = 200\pi (both end faces) instead of the base area πr2\pi r^2⁠.
    Working backwards from each option is a quick, independent check: only (A) reproduces the given 314 m³/h. With the exact π\pi the rate is 0.99950.9995 m/h, within 0.0010.001 of (A) and at least 0.10.1 away from every other option, so the choice does not depend on how π\pi is rounded.
    Rate of change of a quantity, Section 6.2. Lesson: rate of change of quantities
Result

V=100πhV = 100\pi h⁠, so dhdt=314100π≈1\frac{dh}{dt} = \frac{314}{100\pi} \approx 1 m/h (exactly 1 with π=3.14\pi = 3.14⁠). The answer is (A) 1 m/h. (The answer key in this reprint has no entry for this question.)

Check it with numbers

In one hour 314 m³ of wheat spreads over a base of area π(10)2≈314.16\pi(10)^2 \approx 314.16 m², so it forms a layer about 314314.16≈0.9995\frac{314}{314.16} \approx 0.9995 m thick: the depth goes up by about 1 m every hour ✓. Conversely, a depth rate of 0.99949 m/h corresponds to 100π×0.99949≈314.0100\pi \times 0.99949 \approx 314.0 m³/h ✓.

Common mistakes
  1. Using the circumference 2πr2\pi r instead of the area πr2\pi r^2⁠: 3142π⋅10≈5\frac{314}{2\pi\cdot 10} \approx 5 m/h, not even an option.
  2. Using rr instead of r2r^2: 31410π≈10\frac{314}{10\pi} \approx 10 m/h.
  3. Thinking the rate changes as the tank fills. For a cylinder the base area is constant, so the depth rises at a constant rate; compare a cone, where it does not (Example 31).
What to take away
  1. Constant cross-section AA⁠: dhdt=1A dVdt\frac{dh}{dt} = \frac{1}{A}\,\frac{dV}{dt}⁠.
  2. Check an MCQ by working backwards from each option to the given data.
  3. Numbers like 314 or 3.14 in a question are a hint that π≈3.14\pi \approx 3.14 is meant to cancel.

Every fact used, and where it comes from

FactUsed inSource
Volume of a cylinder πr2h\pi r^2 hStep 1Mensuration (Basics: area and volume formulas)
Chain rule for related ratesStep 2Section 6.2; chain rule Theorem 4, Section 5.3.1 (Basics: related rates)
Rate of change as a derivative; sign meaningSteps 3, 4Section 6.2 (Lesson: rate of change of quantities)
All valuesSteps 3, 4, CheckRecomputed in the site's verification script
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