Rate of change of quantities

What is a rate of change?

In plain words: the derivative measures how fast one quantity changes as another changes, as ds/dt is velocity when s denotes displacement (and speed when it denotes distance travelled), in Class 11’s velocity at an instant. Its sign tells the direction: a positive derivative indicates increase and a negative derivative indicates decrease; an increasing function can still have zero derivative at an isolated point (NCERT’s Note, page 149).

The units of dy/dx are the units of y per unit of x: cm² per cm for the area of a circle against its radius, cm²/s for the area against time.

Rate with respect to another quantity

NCERT Example 1 (page 148): the area of a circle is A = πr², so dA/dr = 2πr, which is 10π when r = 5 cm. The area grows by about 10π cm² for each extra centimetre of radius at that moment.

NCERT slip (page 148): Example 1 asks for the rate “per second with respect to its radius” and gives the answer in cm²/s. A rate with respect to the radius is in cm² per cm, as NCERT’s own answers to Exercise 6.1, Q1 give it; no time is involved. The number 10π is right. In an exam, give a rate with respect to the radius in cm² per cm.

Related rates: linking through time

When several quantities change with time, differentiate the equation linking them with respect to t, using the chain rule. NCERT Example 2 (page 148): a cube’s volume V = x³ grows at 9 cm³/s. Then 9 = 3x² dx/dt, so dx/dt = 3/x², and the surface area S = 6x² changes at dS/dt = 12x × 3/x² = 36/x, which is 3.6 cm²/s when x = 10 cm.

NCERT Example 3 (page 149): ripples spread at dr/dt = 4 cm/s, so the enclosed area changes at dA/dt = 2πr dr/dt = 2π(10)(4) = 80π cm²/s when r = 10 cm.

NCERT Example 4 (page 149): a rectangle’s length x falls at 3 cm/min (dx/dt = −3) while its width y grows at 2 cm/min. With x = 10 cm and y = 6 cm, the perimeter P = 2(x + y) changes at 2(−3 + 2) = −2 cm/min, and the area A = xy at y dx/dt + x dy/dt = −18 + 20 = 2 cm²/min.

Marginal cost and marginal revenue

In economics, the marginal cost is the rate of change of total cost with respect to the output, dC/dx, and the marginal revenue is dR/dx (NCERT Examples 5 and 6, page 150).

NCERT Example 5: C(x) = 0.005x³ − 0.02x² + 30x + 5000 gives MC = 0.015x² − 0.04x + 30. At x = 3, MC = 0.135 − 0.12 + 30 = 30.015, about ₹30.02. NCERT Example 6: R(x) = 3x² + 36x + 5 gives MR = 6x + 36, which is ₹66 at x = 5.

Solved examples

Common mistakes to avoid

Quick revision summary

• If y = f(x), dy/dx is the rate of change of y with respect to x; a positive derivative indicates increase, while an increasing function may also have isolated zero derivatives.
• For quantities changing with time, differentiate the linking equation with respect to t by the chain rule, then substitute the values: for a circle, dA/dt = 2πr dr/dt.
• Marginal cost is dC/dx and marginal revenue is dR/dx.

Questions students ask

What is the difference between related rates and implicit differentiation?

Related rates problems use implicit differentiation with respect to time: each changing quantity is a function of t, so each term picks up a rate such as dx/dt, while implicit differentiation with respect to x gives dy/dx. For a rectangle of fixed area xy = 36 cm², differentiating with respect to t gives y dx/dt + x dy/dt = 0, but with respect to x it gives dy/dx = −y/x. The chain rule links the two, as dy/dt is dy/dx times dx/dt: when x = 9 cm, y = 4 cm and dx/dt = 3 cm/s, dy/dt = −(4/9) × 3, that is −4/3 cm/s.

What is the difference between related rates and optimisation problems?

Related rates and optimisation use derivatives differently: related rates asks how fast a quantity changes at an instant, and the time derivative is the answer, while optimisation asks for its largest or smallest value, which can occur only where the derivative is 0 or undefined, or at an end point. If a square’s side x grows at 0.5 m/s, dA/dt = 2x dx/dt is 7 m²/s when x = 7 m. If 20 m of fencing makes three sides of a rectangle by a wall, two sides x m long, the area x(20 − 2x) has derivative 20 − 4x, zero at x = 5, and the second derivative −4 < 0 makes 50 m² the largest area.

Why is profit greatest where marginal revenue equals marginal cost?

Profit is greatest where marginal revenue equals marginal cost because the profit P(x) = R(x) − C(x) has derivative P′(x) = MR − MC, which must be 0 at a maximum inside the range of outputs, as long as R and C are differentiable there. The condition is necessary but not sufficient, and P″(x) < 0, that is MC increasing faster than MR (C″ > R″), confirms a local maximum. A global maximum also requires comparison over the allowed outputs. With revenue 50x and cost 0.1x² + 10x + 300 rupees, MR = 50 and MC = 0.2x + 10 are equal at x = 200, and P″ = −0.2 < 0, so 200 units give the largest profit, ₹3700. Below 200 units MR > MC, so each extra unit adds to the profit, and beyond 200 each extra unit takes some away.

What is the difference between related rates and differential equations?

In a related rates problem the equation linking the quantities is known, and differentiating it gives one rate from another at an instant, while a differential equation relates an unknown function to its derivatives, and solving it, usually by integrating, finds the function itself. If a circle’s radius grows at dr/dt = 2 cm/s, related rates gives the area’s rate dA/dt = 2πr dr/dt, here 4πr, which is 28π cm²/s when r = 7 cm. Solving the differential equation dr/dt = 2 with r = 1 cm at t = 0 instead gives the radius at every time, r = 2t + 1 (in cm, with t in seconds).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. The volume of a cube increases at 8 cm³/s. How fast is the surface area increasing when an edge is 12 cm? (NCERT Exercise 6.1, Q2, page 150)

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   3x² dx/dt = 8, so dx/dt = 8/(3x²), and dS/dt = 12x dx/dt = 32/x = 32/12 = 8/3 cm²/s.

2. Q2. A length x decreases at 5 cm/min and a width y increases at 4 cm/min. When x = 8 cm and y = 6 cm, find the rates of change of the perimeter and of the area (NCERT Exercise 6.1, Q7, page 151).

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   dP/dt = 2(−5 + 4) = −2 cm/min. dA/dt = y dx/dt + x dy/dt = 6(−5) + 8(4) = 2 cm²/min.

3. Q3. The radius of an air bubble increases at ½ cm/s. At what rate is its volume increasing when the radius is 1 cm? (NCERT Exercise 6.1, Q12, page 151)

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   dV/dt = 4πr² dr/dt = 4π(1)(½) = 2π cm³/s.

4. Q4. R(x) = 3x² + 36x + 5. The marginal revenue at x = 15 is (A) 116; (B) 96; (C) 90; (D) 126 (NCERT Exercise 6.1, Q18, page 152).

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   MR = 6x + 36 = 90 + 36 = 126: (D).