Practice Questions on Applications of Derivatives

In this article, we are going to see solved questions and also practice questions for a better understanding of the concept of the applications of derivatives.

Identifying Increasing and Decreasing Functions

Derivatives are employed to determine if a given function is increasing, decreasing, or constant, especially in graphical representations. If a function f is continuous within an interval [p, q] and differentiable in the open interval (p, q), then:

f is increasing in [p, q] if f'(x) > 0 for all x ∈ (p, q)

f is decreasing in [p, q] if f'(x) < 0 for all x ∈ (p, q)

f is a constant function in [p, q] if f'(x) = 0 for all x ∈ (p, q)

Tangent and Normal Lines

Tangent lines touch a curve at a single point without crossing it, while normal lines are perpendicular to tangents.

For a point P(x1, y1) on the curve, the equation of the tangent is given as:

y – y1 = f'(x1)(x – x1)

The equation of the normal is:

y – y1 = [-1 / f'(x1)](x – x1)

Maxima and Minima

Derivatives are pivotal in determining the highest and lowest points on a curve, also known as maxima and minima. These points are vital in understanding the turning points of a function.

Monotonicity

Functions can be classified as monotonic if they are either continuously increasing or decreasing across their entire domain. Functions that exhibit both increasing and decreasing behavior are considered non-monotonic.

Approximation or Finding Approximate Values

Derivatives are instrumental in estimating very small variations or changes in a quantity. This involves using the delta symbol (△) to represent approximate values.

Points of Inflection

For continuous functions, points of inflection occur where the second derivative changes sign while the first derivative exists. These points signify changes in the curvature of the curve.

Example 1: Demonstrate that the function f(x) = x³ – 2x² + 2x, where x ∈ Q, is an increasing function on Q.

Solution:

f(x) = x³ – 2x² + 2x

Upon differentiating both sides, we obtain:

f'(x) = 3x² – 4x + 2 > 0

Now, to prove that ?′(?) ≥ 0 for all ? in ?, we can use the discriminant of the quadratic equation 3?²−4?+2:

Δ = (−4)²−4(3)(2) = 16 − 24 = −8

Since the discriminant is negative, the quadratic equation has no real roots. Thus, the derivative ?′(?) is always positive (or zero), implying that ?(?) is an increasing function.

Hence, f is increasing on Q.

Example 2: The tangent line to the curve y = x² − 5x + 5, which is parallel to the line 2y = 4x + 1, also passes through another point. Determine the coordinates of this point.

Solution:

To find the tangent line to the curve y = x² – 5x + 5 that is parallel to the line 2y = 4x + 1, we need to find the point where these lines intersect.

First, let’s rearrange the equation of the given line 2y = 4x + 1 into slope-intercept form:

y = 4/2 x + 1/2

y = 2x + 1/2​

So, the slope of the given line is m = 2.

Now, the tangent line to the curve at any point (x₀,y₀) has a slope equal to the derivative of the curve at that point. Let’s find the derivative of the curve y = x² – 5x + 5:

dy/dx = 2x – 5

For the tangent line to be parallel to the given line, its slope must also be m = 2. Therefore, we set 2x – 5 = 2 and solve for x:

2x – 5 = 2

2x = 2 + 5

2x = 7

x = 7/2

Now, plug x = 7/2 into the original curve equation to find the corresponding y value:

y = (7/2)² – 5(7/2) + 5

y = -1/4

So, the coordinates of the point where the tangent line intersects the curve are (7/2, -1/4).

Example 3: The tangent line to the curve y = e^2x − x² which passes through the point (1, e) also goes through another point. Find this additional point.

Solution:

Tangent line passes through the point (1, e). the slope of the tangent line at x = 1 to be 2e² – 2.

Now, let’s find the equation of the tangent line using the point-slope form:

y – e = (2e² – 2)(x – 1)

Now, we need to find another point on this line. To do that, we can choose another value of x and then find the corresponding y.

Let’s take x = 2, for instance:

y – e = (2e² – 2)(2 – 1)

y – e = 2e² – 2

y = 2e² – 2 + e

y = 2e² + e – 2

Thus, the additional point through which the tangent line passes is (2, 2e² + e – 2).

Example 4: Show that the function [Tex]g(x) = \frac{x^4}{4} – \frac{x^3}{3} + \frac{x^2}{2} – x[/Tex], where x ? Q, is increasing on Q.

Solution:

To prove that g(x) is increasing on Q, we need to demonstrate that its derivative is non-negative over this domain.

First, let’s find the derivative of  g(x):

g'(x) = x³ – x² + x – 1

Now, to prove that  g'(x) >= 0 for all  x in Q, we can use the polynomial itself and factor it:

g'(x) = x³ – x² + x – 1 = x²(x – 1) + 1(x – 1) = (x² + 1)(x – 1)

The quadratic factor  x² + 1 is always positive for real  x, and the linear factor  x – 1 is also positive for  x > 1 and negative for  x < 1.

Since both factors are positive for  x > 1 and both factors are negative for  x < 1,  g'(x) is positive for  x > 1 and negative for  x < 1.

Therefore,  g'(x) is non-negative for all  x in Q, implying that g(x) is an increasing function on Q.

Example 5: A rectangular garden is to be constructed using 100 meters of fencing material. Find the dimensions of the garden that will maximize its area.

Solution:

Let x and y be the length and width of the rectangular garden, respectively.

Given that the perimeter of the rectangular garden is 100 meters,

2x + 2y = 100

x + y = 50

Now, we want to maximize the area A of the garden, given by:

A = xy

We can rewrite y in terms of x using the equation  x + y = 50, which gives us  y = 50 – x.

Substitute this expression for y into the equation for area:

A = x(50 – x)

A = 50x – x²

Now, to find the critical points where  A’ = 0 we take the derivative of A with respect to x and set it equal to zero:

A'(x) = 50 – 2x

50 – 2x = 0

2x = 50

x = 25

So,  x = 25 meters. 

Example 6: Show that the function f(x) = x – sin x is increasing on R?

Solution:

Given: f(x) = x – sin x

Let’s calculate f'(x).

⇒ f'(x) = 1 – cos x

As we know that, – 1 ≤ cos x ≤ 1 ∀ x ∈ R

⇒ f'(x) = 1 – cos x ≥ 0 ∀ x ∈ R

As we know that for an increasing function say f(x) we have f'(x) ≥ 0

Hence, the given function f(x) is an increasing function on R.

Example 7: A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at a rate of 0.05 cm per second. Find the rate at which its area is increasing if the radius is 3.2 cm.

Solution:

Let us assume that “r” be the radius of the given disc and “A” be the area, then the area is given as:

A = πr²

By using the chain rule,

Then dA/dt = 2πr(dr/dt)

Given: The approximate rate of increase of radius = (dr/dt) = 0.05 cm per second

Hence, the approximate rate of increase in area is:

dA/dt = 2πr[(dr/dt) at r = 3.2 cm

= 2π (3.2) (0.05)

= 0.320π cm² per second.

Therefore, when r= 3.2 cm, then the area is increasing at a rate of 0.320π cm²/second.

Example 8: What is the equation of the normal to the curve y = sin x at (0, 0)?

(a)x =0 (b) y=0 (c)x + y = 0 (d)x – y = 0

Solution:

A correct answer is an option (c)

Explanation:

Given that, y = sin x

Hence, dy/dx = cos x

Thus, the slope of the normal = (-1/cos x)

at x =0, slope of the normal = -1

Therefore, the equation of the normal is y-0 = -1(x-0) or x+y=0

Hence, the correct solution is option c.

Example 9 The total revenue received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5 in rupees. Find the marginal revenue when x = 5, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant.

Solution:

To find the marginal revenue, we need to find the derivative of the revenue function R(x) with respect to the number of items sold x, and then evaluate it at the given value of x = 5.

Given the revenue function:

R(x) = 3x² + 36x + 5

We differentiate R(x) with respect to x to find the marginal revenue function MR(x):

MR(x) = dR/dx

MR(x) = d/dx (3x² + 36x + 5)

MR(x) = 6x + 36

Now, we evaluate MR(x) at x = 5:

MR(5) = 6(5) + 36

MR(5) = 30 + 36

MR(5) = 66

So, when x = 5, the marginal revenue is 66 rupees. This means that for each additional unit sold when x = 5, the revenue will increase by 66 rupees.

Example 10 Show that the function f(x) = x³ – 3x² + 6x – 100 is increasing on R.

Solution:

Given, f(x) = x³ – 3x² + 6x – 100

On differentiating both sides w.r.t. x, we get

f'(x) = 3x² – 6x + 6

the discriminant of the quadratic equation is:

Δ = (−6)2 − 4(3)(6) = 36 − 72 = −36

Since the discriminant is negative, the quadratic equation has no real roots.

Thus, the derivative ?′(?) is always positive (or zero), implying that ?(?) is an increasing function on ?.

Example 1: Find the rate of change of the area of a circle with respect to its radius r when r = 6 cm.

Example 2: Find the equation of a tangent to the curve y=x4–6×3+13×2–10x+5 at the point (1, 3).

Example 3: Amongst all the pairs of positive numbers with sum 24, find those whose product is maximum.

Example 4: Find the Stationary point of the function f(x)=x2−x+6.

Example 5: An edge of a variable cube is increasing at the rate of 5 cm/sec. How fast is the volume of the cube increasing when the edge is 10 cm long?

Example 6: The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate 4 cm/minute. When x = 8 cm and y = 6 cm then find the rate of change of the area of the rectangle.

Example 7: A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area is increasing when its radius is 5 cm? (Take π = 3.14).

Example 8: A stone is dropped into a quite pond and the waves moves in circles. If the radius of the circular wave increases at the rate of 8 cm/sec, find the rate of increase in its area at the instant when its radius is 6 cm?

Example 9: Find the rate of change of the area of a circle with respect to its radius r when r = 6 cm.

Example 10: If radius of circle is increasing at rate 0.5 cm/sec what is the rate of increase of its circumference?

Example 11: Find the Stationary point of the function f(x)=x2−x+8.

Example 12: Find the rate of change of the area of a circle with respect to its radius r when r = 9 cm.

Related Article:

• Derivative of a Function
• Derivative Rules
• Application of Derivatives
• Differentiation

What are the applications of derivatives in real life?

Derivatives are used in various real-life scenarios such as calculating rates of change, optimizing functions in economics and engineering, predicting stock market trends, and modeling physical phenomena like motion and growth.

How are derivatives used in economics?

In economics, derivatives are utilized to analyze and optimize production functions, determine marginal cost and revenue, evaluate consumer and producer surplus, and understand elasticity of demand and supply.

What role do derivatives play in physics and engineering?

Derivatives are fundamental in physics and engineering for modeling and analyzing dynamic systems, such as motion, fluid flow, heat transfer, and electrical circuits. They help in understanding rates of change, optimization, and stability analysis.

Can derivatives be applied in biology and medicine?

Yes, derivatives are applied in biology and medicine to model population growth, study enzyme kinetics, analyze drug dosage and pharmacokinetics, and understand biological processes like gene expression and signal transduction.

How are derivatives used in finance and investment?

In finance, derivatives are employed for risk management, portfolio optimization, option pricing, hedging strategies, and assessing financial instruments’ sensitivity to market changes, aiding investors and institutions in decision-making.