Sample Problems on Applications of Derivatives

Problem 1: Find the equation of the tangent and the normal to the circle having equation x² + y² = a² at a point (3, 6).

Solution:

Given, Equation of circle = x² + y² = a².

Differentiating the above equation with respect to x,

2 × x + 2 × y dy/dx = 0

⇒ dy/dx = -(2 × x) / (2 × y)

⇒ dy/dx = -(x / y)

Equation of tangent: (Y-y) = (dy/dx) × (X – x)

⇒ (Y – y) = -(x / y) ×  (X – x)

⇒ (Y × y) – y = -(X × x) + x² [Multiplying left and right side by y]

⇒ (Y × y) + (X × x) = x² + y²

⇒ (Y × y) + (X × x) = a²

Putting x = 3 and y = 6, 

(Y × 6) + (X × 3) = a², this is the required equation.

Equation of normal: (Y – y) = (-dx/dy) × (X – x)

⇒ (Y – y) = (y / x) × (X – x), -dx/dy = y / x

⇒ (Y × x) – y × x = (X × y) – y × x

⇒ (Y × x) – (X × y) = 0   

Putting x = 3 and y = 6,

(Y × 3) – (X × 6) = 0, this is the required equation.                                 

Problem 2: Find the equation of the tangent to the ellipse having equation (x₂ / a₂) + (y₂ / b₂) = 1 at a point (x₁, y₁).

Solution:

Given, Equation of ellipse = (x² / a²) + (y²/ b²) = 1

Differentiating the  above the equation with respect to x,

(2 × x) / a² + ((2 × y) / b² ) × (dy/dx) = 0

⇒ dy/dx = (-(2 × x) / a²) / ((2 × y) / b²)

⇒ dy/dx = (- x × b²) / (y × a²)

Now, dy/dx at (x₁, y₁) = (-x₁ × b²) / (y₁ × a²)

Equation of tangent: (Y – y₁) = (dy/dx) × (X – x₁)

(Y – y₁) = ((-x₁ × b²) / (y₁ × a²)) × (X – x₁)

⇒ (Y × y₁ × a²) –  (y₁² × a²) = (- X × x₁ × b²) + (x₁² × b²)

Dividing both sides by (a² × b²),

((Y × y₁) / b²) – (y₁² / b²) = -(( X × x₁) / a²) + (x₁² / a₂)

⇒ ((X × x₁) / a²) + ((Y × y₁) / b²) = (x₁² / a²) + (y₁² / b²)

⇒ ((X × x₁) / a²) + ((Y × y₁) / b²) = 1, this is the required equation.     

⇒ (x₁² / a²) + (y₁² / b²) = 1

Problem 3: Find the equation of normal to a curve having equation x²+ y² – 2 × x – 10 × y  + 16 = 0 at point (2, 2).

Solution:                      

Given, Equation of curve: x² + y² – 2 × x – 10 × y + 16 = 0

Differentiating the equation with respect to x, 

2 × x + 2 × y – 2 – (10 × dy/dx) = 0

dy/dx = (- (2 × x) – (2 × y) + 2) / -10

Putting x = 2 and y = 2,

dy/dx = 6/10 = 3 / 5

⇒ -dx/dy = -(5/3)

Equation of  normal: (Y – y) = (-dx/dy) × (X – x)

⇒ (Y – 2) = -(5/3) × (X – 2)

⇒ (3 × Y)  – 6 = (- 5 × X) + 10

⇒ (3 × Y) + (5 × X) = 16, this is the required equation.

Problem 4: Find the equation of the tangent to the parabola having equation y² = 4 × a × x at the point (x₁, y₁).

Solution:

Given, Equation of parabola: y² = 4 × a × x

Differentiating the equation with respect to x,

2 × y × dy/dx = 4 × a

⇒ dy/dx = (4 × a) / (2 × y)

⇒ dy/dx = (4 × a) / (2 × y₁) at (x₁,y₁)

Equation of tangent at (x₁, y₁) is given by: (Y – y₁) = (dy/dx) × (X – x₁)

(Y – y₁) = ((4 × a) / (2 × y₁)) × (X – x₁)

⇒ Yy₁ – y₁² = 2aX – 2ax₁

⇒ (Y × y₁) – (2 × a × X) – (2 × a × x₁) = y₁² – (4 × a × x₁), subtracting 2 × a × x₁ from both sides

(Y × y₁) = -2 × a × (X + x₁) ,this is the required equation, y₁²– (4 × a × x₁) = 0

Problem 5: Find the equation of the tangent to the curve having equation 4 × x² + 9 × y² = 72 at point (3, 2).

Solution:

Given, Equation of curve: 4 × x² + 9 × y² = 72

Differentiating the equation with respect to x,

8 × x + 18 × y × dy/dx = 0

dy/dx = (-8 × x) / (18 × y)

Putting x = 3 and y = 2,

dy/dx = -24 / 36 = -2 / 3

Equation of tangent: (Y – 2) = (- 2 / 3) × (X – 3)

(3 × Y) – 6 = (- 2 × X) + 6

(3 × Y) + (2 × X) = 12, this is the required equation. 

Note: Trick to write the equation of a tangent to a curve at the point (x₁,y₁)  

1. Replace x² and y² in the equation of curve by (x × x₁) and (y × y₁) respectively. 
2. Replace x and y by (x + x₁) / 2 and (y + y₁) / 2 respectively.
3. Replace (x × y) by ((x × y₁) + (y × x₁)) / 2
4. Constants remain unchanged.

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. This makes Derivatives very useful in various fields. For example, derivatives help in understanding motion, growth, and change in physical, economic, and engineering systems. They are used to find rates of change, slopes of curves, and to solve optimization problems. By understanding derivatives, we can predict how things will change and make better decisions based on this information.