Derivatives as Rate of Change

Derivatives are fundamental to differential calculus. They describe how a function behaves, such as increasing or decreasing. Suppose we have two quantities, x and y, that vary together and are related by the function y = f(x). The derivative of this function, denoted as\frac{dy}{dx}, represents the rate of change of y with respect to x. This tells us how y changes as x changes.

For example: Find the rate of change of volume of a cube whose sides are increasing at the rate of 2 m/s. 

Solution: 

Let's say the length of the side of cube is "a". The volume of cube is given by, V = a³. 

\frac{dV}{da} = \frac{d(a^3)}{da}

⇒ \frac{dV}{dt} = 3a^2\frac{da}{dt}

⇒\frac{dV}{dt} = 3a^2(2)

⇒\frac{dV}{dt} = 6a² m³/s. 

Increasing and Decreasing Functions 

Derivatives are also used in finding out whether the function is increasing or decreasing or none of them. The figure given below shows the function f(x) = x². 

Notice in the figure, the function is decreasing in the interval (-∞, 0) and increasing in the interval (0,∞). 

In an interval I contained in the domain of real valued function “f”. Then, f is said to be, 

• Increasing on I, if x₁ < x₂ in I ⇒ f(x₁) ≤ f(x₂) for all x₁, x₂ ∈ I.
• Strictly Increasing on I, if x₁ < x₂ in I ⇒ f(x₁) < f(x₂) for all x₁, x₂ ∈ I.

• Decreasing on I, if x₁ < x₂ in I ⇒ f(x₁) ≥ f(x₂) for all x₁, x₂ ∈ I.
• Strictly Decreasing on I, if x₁ < x₂ in I ⇒ f(x₁) > f(x₂) for all x₁, x₂ ∈ I.

Now we know the definitions for increasing and decreasing functions. Let's see how to recognise a function that is increasing or decreasing in an interval. 

Let's say f is continuous on [a, b] and differentiable on the open interval (a, b). Then, 

1. f is increasing in (a, b) if f'(x) > 0 in the interval [a, b].
2. f is decreasing in (a, b) if f'(x) < 0 in the given interval.
3. f is constant if f'(x) = 0.

Sample Problems on Derivatives as Rate of Change

Question 1: Let's say we have a circle whose radius is increasing. Find the rate of change of area with radius when r = 4cm. 

Solution: 

Let's say "A" is the area of the circle and “r” be the radius of the circle.

A = πr²

Differentiating it with respect to radius. 

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

⇒ \frac{dA}{dr} = 2\pi r

At r = 4. 

\frac{dA}{dr} = 8\pi

Question 2: Let's say we have a rectangle whose sides are changing every second. The length is increasing at the rate of 3 m/s while the breadth is increasing at 8 m/s. Calculate the rate at which the area of the rectangle is increasing when the length = 8m and the breadth = 5m. 

Solution: 

Let, x be the length of the rectangle and y be the breadth of rectangle. 

\frac{dx}{dt} = 3 And \frac{dy}{dt} = 8

The area of rectangle is given by, 

A = xy

Differentiating the equation w.r.t time. 

\frac{dA}{dt} = \frac{d(xy)}{dt}

⇒ \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}

⇒ \frac{dA}{dt} = x\frac{dy}{dt} + y\frac{dx}{dt}

⇒\frac{dA}{dt} = 8(\frac{dy}{dt}) + 5(\frac{dx}{dt})

⇒ \frac{dA}{dt} = 8(8) + 5(3)

⇒ \frac{dA}{dt} = 64 + 15

Question 3: For the given curve, find the points where the value of the rate of change of y is zero: y = x² + x

Solution: 

y = x² + x

\frac{dy}{dx}  = \frac{d(x^2 + x)}{dx}

⇒\frac{dy}{dx}  = 2x + 1

This rate of change must be zero, 

2x + 1 = 0 

⇒ x = \frac{-1}{2}

Thus, at x = \frac{-1}{2} the rate of change is zero. 

Question 4: Prove that the function discussed above, f(x) = x² is increasing in the interval (0, ∞). 

Solution: 

According to above definition, a function in increasing in any interval if its derivative is greater than zero in that interval. 

f(x) = x²

Differentiating with respect to x, 

f'(x) = 2x 

For the given interval (0,∞) f'(x) > 0. 

Thus, the function is increasing in the given interval. 

Question 5: Find the intervals where the function f(x) = x² + 5x + 6 is increasing or decreasing. 

Solution: 

Given f(x) = x² + 5x + 6

f'(x) = 2x + 5

We need to study the sign of the derivative to find the intervals where this function is increasing or decreasing. 

f'(x) < 0 

⇒ 2x + 5 < 0 

⇒ x < \frac{-5}{2}

f'(x) > 0 

⇒ 2x + 5 > 0 

⇒ x > \frac{-5}{2}

Thus, the function is increasing in (\frac{-5}{2}, ∞) and is decreasing in (-∞, \frac{-5}{2}). 

Practice Problems on Derivatives as Rate of Change

Question 1: If the radius of a circle is increasing at a uniform rate of 2 cm/sec, find the rate of increasing of area of circle, at the instant when the radius is 20 cm.

Question 2: The percentage error in calculating the volume of a cubical box if an error of 1% is made in measuring the length of edges of the cube is?

Question 3: if s= \frac{1}{2}t^{3}-6t , Calculate the acceleration at the time when the velocity becomes zero.

Question 4: Prove that the function f(x) = x/log x increases on the interval (e, ∞).

Question 5: Find the intervals where the function y = \frac{3}{2}x^{64} -3x^2 + 1 is increasing or decreasing.