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Rate of change

A derivative dydx\dfrac{dy}{dx} is the instantaneous rate at which yy changes as xx changes. When xx is time, dydt\dfrac{dy}{dt} is a velocity, growth rate, or flow rate. When two quantities both depend on time and are related by an equation, the chain rule connects their rates , the related rates problem.

Single variable rate

If y=f(x)y = f(x) and xx varies with time, then

dydt=f(x)dxdt.\frac{dy}{dt} = f'(x) \cdot \frac{dx}{dt}.

So f(x)f'(x) is the conversion factor between rates.

Worked example: a spreading oil slick

An oil slick is a growing circle with radius rr increasing at 22 cm/s. Find the rate at which the area A=πr2A = \pi r^2 is growing when r=5r = 5.

dAdt=2πrdrdt=2π52=20π\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt} = 2\pi \cdot 5 \cdot 2 = 20 \pi cm2^2/s.

  1. Identify all variables, sketch if helpful.
  2. Write the equation relating the variables.
  3. Differentiate both sides with respect to time tt, using the chain rule.
  4. Substitute the given values to find the desired rate.

Worked example: a ladder sliding down a wall

A 1010-m ladder leans against a wall. The bottom slides away from the wall at 0.50.5 m/s. How fast is the top sliding down when the bottom is 66 m from the wall?

Let xx = horizontal distance, yy = vertical distance. x2+y2=100x^2 + y^2 = 100.

Differentiate: 2xx˙+2yy˙=02x\dot x + 2y\dot y = 0, so y˙=xx˙y\dot y = -\dfrac{x \dot x}{y}.

At x=6x = 6: y=8y = 8. y˙=60.58=3/8\dot y = -\dfrac{6 \cdot 0.5}{8} = -3/8 m/s.

The top is sliding down at 3/83/8 m/s.

Worked examples

Example 1. A balloon's volume V=43πr3V = \dfrac{4}{3}\pi r^3 increases at 2020 cm3^3/s. Find the rate of change of radius when r=2r = 2 cm.

dVdt=4πr2drdt\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}, so 20=4π(4)drdt20 = 4\pi(4)\dfrac{dr}{dt}, giving drdt=54π\dfrac{dr}{dt} = \dfrac{5}{4\pi} cm/s.

Example 2. The radius of a circle is increasing at 33 cm/s. Find the rate of change of its area when r=10r = 10 cm.

dAdt=2πrdrdt=60π\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt} = 60\pi cm2^2/s.

Example 3. Water is poured into a conical tank (apex down) of height 1010 m and base radius 55 m at 22 m3^3/min. How fast is the water level rising when the depth is 44 m?

By similar triangles, water radius r=h/2r = h/2. Volume V=13πr2h=πh312V = \tfrac{1}{3}\pi r^2 h = \tfrac{\pi h^3}{12}. Differentiate: dVdt=πh24dhdt\dfrac{dV}{dt} = \tfrac{\pi h^2}{4}\dfrac{dh}{dt}. Substitute: 2=π164h˙2 = \tfrac{\pi \cdot 16}{4}\dot h, so h˙=1/(2π)\dot h = 1/(2\pi) m/min.

Example 4. A particle moves along y=x2+1y = x^2 + 1. Find the rate of change of yy when x=2x = 2 and dxdt=5\dfrac{dx}{dt} = 5.

dydt=2xdxdt=225=20\dfrac{dy}{dt} = 2x \dfrac{dx}{dt} = 2 \cdot 2 \cdot 5 = 20.

Example 5. The side of a square is increasing at 0.50.5 cm/s. Find the rate of increase of the diagonal when the side is 44 cm.

Diagonal d=s2d = s\sqrt 2. dddt=2dsdt=22=12\dfrac{dd}{dt} = \sqrt 2 \dfrac{ds}{dt} = \dfrac{\sqrt 2}{2} = \dfrac{1}{\sqrt 2} cm/s. (Independent of side length.)

Example 6. Total revenue from sale of xx units is R(x)=13x2+26x+15R(x) = 13 x^2 + 26 x + 15. Find the marginal revenue at x=7x = 7.

Marginal revenue =R(x)=26x+26= R'(x) = 26 x + 26. At x=7x = 7: 267+26=20826 \cdot 7 + 26 = 208.

Try it yourself

  1. The radius of a sphere is increasing at 0.50.5 cm/s. Find the rate of change of volume when r=4r = 4 cm.
  2. A man 22 m tall walks away from a 66-m lamp post at 1.51.5 m/s. Find the rate at which his shadow lengthens.
  3. The side of an equilateral triangle increases at 22 cm/s. Find the rate of change of area when the side is 1010 cm.
  4. A cylindrical tank of radius 33 m has water rising at 0.50.5 m/min. Find the volume flow rate.
  5. The radius of a circular ripple in water increases at 44 cm/s. Find the rate of growth of the area when r=6r = 6 cm.
  6. Total cost C(x)=0.007x30.003x2+15x+4000C(x) = 0.007 x^3 - 0.003 x^2 + 15 x + 4000. Find the marginal cost at x=17x = 17.
  7. Find the rate of change of the surface area of a sphere when the radius is 55 and growing at 22 cm/s.
  8. A particle moves so that s(t)=t36t2+9t+5s(t) = t^3 - 6t^2 + 9t + 5. Find the velocity and acceleration at t=2t = 2.
  9. Volume of a cube increases at 99 cm3^3/s. How fast is the surface area increasing when the edge is 1010 cm?
  10. The volume of a sphere is increasing at 8π8\pi cm3^3/s. Find the rate of change of its surface area when r=5r = 5.
  11. A kite is at height 8080 m flying horizontally at 55 m/s. How fast is the string being released when its length is 100100 m?
  12. The radius of a balloon is decreasing at 11 cm/s. Find the rate of change of volume when r=3r = 3 cm.
  13. A ladder of length 1313 m leans against a wall. Its foot is being pulled away at 22 cm/s. Find the rate at which the top slides down when the foot is 55 m from the wall.
  14. A trough is in the shape of an inverted prism. Water flows in at 22 m3^3/min. Find the rate of rise of water when depth is 11 m if the cross-section is an equilateral triangle of side 22 m.

Pitfalls / Tricks

  • Identify variables and rates first, write the constraint equation, then differentiate.
  • Chain rule with respect to time: every variable that depends on tt contributes a d/dtd/dt factor.
  • Be careful with signs: rates can be negative (e.g., a ladder top sliding down).
  • Always substitute given values after differentiating, never before.
  • Units must be consistent , convert if necessary.

Next, monotonicity from the sign of ff'.

Practice quiz

Quick check on this topic.

Quiz
Quick check — Rate of change
6 questions · pick the best answer
Q1

If s=t33ts=t^3-3t, the velocity at t=2t=2 is:

Q2

The area of a circle is increasing at 4 cm2^2/s. When r=2r=2 cm, drdt\frac{dr}{dt} is:

Q3

A ladder 5 m long leans against a wall. If the foot is pulled away at 1 m/s, the top slides down when foot is 3 m from wall at:

Q4

If V=43πr3V=\frac{4}{3}\pi r^3, then dVdr\frac{dV}{dr} at r=3r=3 is:

Q5

A particle moves along x=t24t+3x=t^2-4t+3. Its velocity is zero at:

Q6

Side of a square increases at 0.5 cm/s. When side is 4 cm, area increases at: