Popular Courses
Beginner – Linear Regression
Introduction
Beginner – Decision Trees
Introduction
Beginner – Neural Networks
Introduction

Sign in to Curvingo

or

New Account     Reset Password     

Plans

Details
Application of Derivatives
Introduction
Rate of Change of Quantities
Practice: Rates of Change
Increasing and Decreasing Functions
Practice: Increasing and Decreasing Functions
Maxima and Minima
Practice: Maxima and Minima
Miscellaneous Examples
Chapter Review: Mixed Practice
Summary
Curvingo · Made in India

Share

Send us Feedback

Please let us know if you have any feedback and suggestions, or if you find any errors and bugs in our content.

Sorry, your message couldn’t be submitted. Please try again!

Thanks for your feedback!

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

Application of DerivativesRate of Change of Quantities

Reading time: ~30 min

The ball's velocity was your first rate of change in this chapter. In general, suppose a quantity y depends on another quantity x, say y = f(x). Then

\frac{dy}{dx} (or f'(x)) is the rate of change of y with respect to x,

and \left(\frac{dy}{dx}\right)_{x = x_0} (or f'(x_0)) is that rate of change at the particular value x = x_0.

For the ball, the height depends on time, so \frac{dh}{dt} is the rate of change of height with respect to . At t = 1, it is h'(1) = 20 - 10 = m/s.

Very often two quantities change together as time passes: the radius and the area of a spreading ripple, the edge and the volume of a melting ice cube. If x = f(t) and y = g(t), the chain rule links their rates:

\frac{dy}{dx} = \dfrac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt} \neq 0.

So the rate of change of y with respect to x can be found from the rates of change of y and of x with respect to t. Read the other way round, \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}, and that is the form you will use most in this section.

Ripples in the Temple Tank

Riya drops a pebble into the still water of the temple tank. A circular ripple spreads out, its edge moving at a steady 4 cm per second. The radius grows at a constant rate. Does the area inside the ripple grow at a constant rate too?

The blue disc is the water inside the ripple. The orange ring is the strip that the edge sweeps over in the next quarter of a second, and its area is roughly (length of the edge) × (width of the strip).

Drag the radius and complete the missions.

Radius of the ripple: ${r} cm

The ring tells the whole story. Its length is the circumference 2\pi r, so the area grows by about 2\pi r square centimetres for every centimetre the radius grows: \frac{dA}{dr} = 2\pi r. A larger ripple has a longer edge, so the same 4 cm/s sweeps in more water. Two of your missions are Examples 1 and 3 below.

Example 1 : Find the rate of change of the area of a circle with respect to its radius r when r = 5 cm.

Solution : The area of a circle of radius r is A = \pi r^2. So the rate of change of the area with respect to the radius is

\frac{dA}{dr} = \frac{d}{dr}\left(\pi r^2\right) =

At r = 5 cm, \frac{dA}{dr} = 2\pi \times 5 = cm² per cm of radius.

So when the radius is 5 cm, the area grows at 10\pi square centimetres for each extra centimetre of radius. That was your first mission.

Example 2 : The volume of a cube is increasing at the rate of 9 cm³ per second. How fast is its surface area increasing when the length of an edge is 10 cm?

Solution : Let the edge be x cm, the volume V and the surface area S. Then V = x^3 and S = 6x^2, and all three change with time t. The volume grows at 9 cm³/s, so by the chain rule

\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt} = 9

which gives \frac{dx}{dt} = .

Now differentiate the surface area:

\frac{dS}{dt} = 12x \cdot \frac{dx}{dt} = 12x \cdot \frac{3}{x^2} =

At x = 10 cm, \frac{dS}{dt} = \frac{36}{10} = cm²/s.

The edge was never given a speed. We found \frac{dx}{dt} from the volume and then used it for the surface area. That two-step move (differentiate one formula to uncover a hidden rate, then feed it into another) is the heart of every "how fast" problem.

Example 3 : A stone is dropped into a quiet lake, and waves move outwards in circles at 4 cm/s. At the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?

Solution : The area enclosed by a circle of radius r is A = \pi r^2, and the radius grows at \frac{dr}{dt} = 4 cm/s. By the chain rule,

\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} =

= 2\pi \times 10 \times 4 = cm²/s.

That was your second mission in the temple tank. At r = 10 cm the enclosed area grows by 80\pi cm² every second, although the edge still moves at only 4 cm/s.

Example 4 : The length x of a rectangle is decreasing at 3 cm/minute and its width y is increasing at 2 cm/minute. When x = 10 cm and y = 6 cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.

Solution : A decreasing quantity has a negative rate of change, so \frac{dx}{dt} = cm/min, while \frac{dy}{dt} = 2 cm/min.

(a) The perimeter is P = 2(x + y), so

\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-3 + 2) = cm/min.

(b) The area is A = xy, so by the product rule

\frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt} = 10(2) + 6(-3) = cm²/min.

Note: read the sign

At this instant the perimeter is shrinking at 2 cm/min (its rate is negative) while the area is growing at 2 cm²/min. One rectangle, two quantities, opposite directions. Always say which way a quantity is changing, not just how fast.

The Sliding Ladder

A painter's ladder, 5 m long, leans against a wall. Its foot starts to slip and slides away from the wall at a steady 1 m/s, so the top slides down the wall. How fast does the top move? Most people guess "also 1 m/s". Test that guess.

Drag the foot of the ladder along the floor. The readout shows how fast the top is sliding down at that moment.

Let the foot be x m from the wall and the top y m up the wall. The ladder is the hypotenuse of a right triangle, so by Pythagoras x^2 + y^2 = at every instant.

Differentiate both sides with respect to t: 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0, so \frac{dy}{dt} =

At x = 3 m the top is at y = 4 m, and with \frac{dx}{dt} = 1 m/s we get \frac{dy}{dt} = -\frac{3}{4} m/s: the top slides down at 0.75 m/s, just as the readout said. As y shrinks towards 0, the fraction \frac{x}{y} grows without limit, which is why the top seems to crash down at the very end.

Marginal Cost and Marginal Revenue

Rates of change are not only about shapes and speeds. Mrs Kapoor runs a small workshop that makes steel tiffin boxes. In economics, the marginal cost is the rate of change of the total cost with respect to the number of items made; roughly, it is the cost of making one more item at the present level of output. In the same way, the marginal revenue is the rate of change of the total revenue with respect to the number of items sold.

If C(x) is the total cost and R(x) the total revenue for x units, then

marginal cost MC = \frac{dC}{dx} and marginal revenue MR = \frac{dR}{dx}.

Example 5 : The total cost C(x), in rupees, of producing x units of an item is C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000. Find the marginal cost when 3 units are produced.

Solution : MC = \frac{dC}{dx} =

At x = 3: MC = 0.015(9) - 0.04(3) + 30 = 0.135 - 0.12 + 30 = 30.015. So the marginal cost is about : making one more unit at this stage costs a little over ₹30.

Example 6 : The total revenue, in rupees, from selling x units of a product is R(x) = 3x^2 + 36x + 5. Find the marginal revenue when x = 15.

Solution : MR = \frac{dR}{dx} = 6x + 36, so at x = 15, MR = 6(15) + 36 = . The marginal revenue is ₹126.

Speed Round: Rates in a Flash

Ripples, balloons, squares, cubes and shops. Write the formula in your head, differentiate it, then put in the numbers.

Get five right to move on.

Sina