Maharashtra Board Class 12 Maths Part 1 Chapter 4 Applications of Derivatives 4.1 Solutions

Question 1. Find the equations of tangent and normal to the following curves at the given point on it:
(i) y = 3x² - x + 1 at (1, 3)
Answer: Solution:
y = 3x² - x + 1
\( \frac{dy}{dx} = \frac{d}{dx} (3x^2 - x + 1) \)
\( = 3 \times 2x - 1 + 0 \)
\( = 6x - 1 \)
\( \left( \frac{dy}{dx} \right)_{\text{at } (1,3)} = 6(1) - 1 \)
\( = 5 \)
= slope of the tangent at (1, 3).
Therefore, the equation of the tangent at (1, 3) is
y - 3 = 5(x - 1)
Therefore, y - 3 = 5x - 5
Therefore, 5x - y - 2 = 0.
The slope of the normal at (1, 3) = \( \frac{-1}{\left( \frac{dy}{dx} \right)_{\text{at } (1,3)}} = -\frac{1}{5} \)
Therefore, the equation of the normal at (1, 3) is
y - 3 = \( -\frac{1}{5} \)(x - 1)
Therefore, 5y - 15 = -x + 1
Therefore, x + 5y - 16 = 0
Hence, the equations of the tangent and normal are 5x - y - 2 = 0 and x + 5y - 16 = 0 respectively.
In simple words: To find the equations of the tangent and normal, first calculate the derivative of the curve to find the slope of the tangent at the given point. The normal's slope is the negative reciprocal of the tangent's slope. Then use the point-slope form to find both equations.

🎯 Exam Tip: Remember that the slope of the tangent is given by the derivative \( \frac{dy}{dx} \) evaluated at the point, and the slope of the normal is \( -\frac{1}{\left( \frac{dy}{dx} \right)} \). Ensure correct algebraic manipulation for final equations.

Question 1.
(ii) 2x² + 3y² = 5 at (1, 1)
Answer: Solution:
2x² + 3y² = 5
Differentiating both sides w.r.t. x, we get
\( 2 \times 2x + 3 \times 2y \frac{dy}{dx} = 0 \)
Therefore, \( 4x + 6y \frac{dy}{dx} = 0 \)
Therefore, \( 6y \frac{dy}{dx} = -4x \)
Therefore, \( \frac{dy}{dx} = \frac{-4x}{6y} = -\frac{2x}{3y} \)
Therefore, \( \left( \frac{dy}{dx} \right)_{\text{at } (1,1)} = \frac{-2(1)}{3(1)} = -\frac{2}{3} \)
= slope of the tangent at (1, 1)
Therefore, the equation of the tangent at (1, 1) is
y - 1 = \( -\frac{2}{3} \)(x - 1)
Therefore, 3y - 3 = -2x + 2
Therefore, 2x + 3y - 5 = 0.
The slope of normal at (1, 1) = \( \frac{-1}{\left( \frac{dy}{dx} \right)_{\text{at } (1,1)}} = \frac{-1}{\left( -\frac{2}{3} \right)} = \frac{3}{2} \)
Therefore, the equation of the normal at (1, 1) is
y - 1 = \( \frac{3}{2} \)(x - 1)
Therefore, 2y - 2 = 3x - 3
Therefore, 3x - 2y - 1 = 0
Hence, the equations of the tangent and normal are 2x + 3y - 5 = 0 and 3x - 2y - 1 = 0 respectively.
In simple words: For an implicit function, differentiate both sides with respect to x, remembering to apply the chain rule for terms involving y. Substitute the point's coordinates to find the slopes and then form the tangent and normal equations.

🎯 Exam Tip: When differentiating implicit functions, ensure every 'y' term is differentiated using the chain rule, resulting in a \( \frac{dy}{dx} \) factor. Simplify the expression for \( \frac{dy}{dx} \) before substituting the point coordinates.

Question 1.
(iii) x² + y² + xy = 3 at (1, 1)
Answer: Solution:
x² + y² + xy = 3
Differentiating both sides w.r.t. x, we get
\( 2x + 2y \frac{dy}{dx} + x \frac{dy}{dx} + y \frac{d}{dx}(x) = 0 \)
Therefore, \( 2x + 2y \frac{dy}{dx} + x \frac{dy}{dx} + y \times 1 = 0 \)
Therefore, \( (x+2y) \frac{dy}{dx} = -2x-y \)
Therefore, \( \frac{dy}{dx} = \frac{-2x-y}{x+2y} = -\frac{(2x+y)}{(x+2y)} \)
Therefore, \( \left( \frac{dy}{dx} \right)_{\text{at } (1,1)} = -\frac{(2(1)+1)}{(1+2(1))} = -\frac{3}{3} = -1 \)
= slope of the tangent at (1, 1)
the equation of the tangent at (1, 1) is
y - 1 = -1(x - 1)
Therefore, y - 1 = -x + 1
Therefore, x + y = 2
The slope of the normal at (1, 1) = \( \frac{-1}{\left( \frac{dy}{dx} \right)_{\text{at } (1,1)}} = \frac{-1}{-1} = 1 \)
Therefore, the equation of the normal at (1, 1) is y - 1 = 1(x - 1)
Therefore, y - 1 = x - 1
Therefore, x - y = 0
Hence, the equations of tangent and normal are x + y = 2 and x - y = 0 respectively.
In simple words: For curves with implicit relations, differentiate each term with respect to x, applying the product rule for xy and chain rule for y². Solve for \( \frac{dy}{dx} \), find its value at the given point, and then use it to write the tangent and normal equations.

🎯 Exam Tip: Pay close attention to the product rule application for terms like 'xy' during implicit differentiation. Common errors occur when students forget to apply the chain rule to 'y' terms or the product rule correctly.

Question 2. Find the equations of the tangent and normal to the curve y = x² + 5 where the tangent is parallel to the line 4x - y + 1 = 0.
Answer: Solution:
Let P(x₁, y₁) be the point on the curve y = x² + 5 where the tangent is parallel to the line 4x - y + 1 = 0.
Differentiating y = x² + 5 w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx} (x^2 + 5) = 2x + 0 = 2x \)
\( \left( \frac{dy}{dx} \right)_{\text{at } (x_1,y_1)} = 2x_1 \)
Let m₁ = 2x₁
The slope of the line 4x - y + 1 = 0 is \( m_2 = \frac{-4}{-1} = 4 \)
Since, the tangent at P(x₁, y₁) is parallel to the line 4x - y + 1 = 0,
m₁ = m₂
Therefore, 2x₁ = 4
Therefore, x₁ = 2
Since, (x₁, y₁) lies on the curve y = x² + 5, y₁ = x₁² + 5
Therefore, y₁ = (2)² + 5 = 9
Therefore, the coordinates of the point are (2, 9) and the slope of the tangent = m₁ = m₂ = 4.
Therefore, the equation of the tangent at (2, 9) is
y - 9 = 4(x - 2)
Therefore, y - 9 = 4x - 8
Therefore, 4x - y + 1 = 0
Slope of the normal = \( \frac{-1}{m_1} = -\frac{1}{4} \)
Therefore, the equation of the normal at (2, 9) is
y - 9 = \( -\frac{1}{4} \)(x - 2)
Therefore, 4y - 36 = -x + 2
Therefore, x + 4y - 38 = 0
Hence, the equations of tangent and normal are 4x - y + 1 = 0 and x + 4y - 38 = 0 respectively.
In simple words: When the tangent is parallel to a given line, their slopes are equal. First, find the slope of the curve and the slope of the given line. Equate them to find the x-coordinate of the point of tangency, then find the y-coordinate. Finally, use the point and slope to write the tangent and normal equations.

🎯 Exam Tip: Remember that parallel lines have equal slopes. This is the key condition for finding the point of tangency. Ensure you calculate the derivative correctly and accurately solve for the coordinates of the point where the tangent exists.

Question 3. Find the equations of the tangent and normal to the curve y = 3x² - 3x - 5 where the tangent is parallel to the line 3x - y + 1 = 0.
Answer: Solution:
Let P(x₁, y₁) be the point on the curve y = 3x² - 3x - 5 where the tangent is parallel to the line 3x - y + 1 = 0.
Differentiating y = 3x² - 3x - 5 w.r.t. x, we get
\( \frac{dy}{dx} = \frac{d}{dx} (3x^2 - 3x - 5) \)
\( = 3 \times 2x - 3 \times 1 - 0 \)
\( = 6x - 3 \)
Therefore, \( \left( \frac{dy}{dx} \right)_{\text{at } (x_1,y_1)} = 6x_1 - 3 \)
= slope of the tangent at (x₁, y₁)
Let m₁ = 6x₁ - 3
The slope of the line 3x - y + 1 = 0 is \( m_2 = \frac{-3}{-1} = 3 \)
Since, the tangent at P(x₁, y₁) is parallel to the line 3x - y + 1 = 0,
m₁ = m₂
Therefore, 6x₁ - 3 = 3
Therefore, 6x₁ = 6
Therefore, x₁ = 1
Since, (x₁, y₁) lies on the curve y = 3x² - 3x - 5,
y₁ = 3x₁² - 3x₁ - 5, where x₁ = 1
= 3(1)² - 3(1) - 5
= 3 - 3 - 5
= -5
Therefore, the coordinates of the point are (1, -5) and the slope of the tangent = m₁ = m₂ = 3.
Therefore, the equation of the tangent at (1, -5) is
y - (-5) = 3(x - 1)
Therefore, y + 5 = 3x - 3
Therefore, 3x - y - 8 = 0
Slope of the normal = \( \frac{-1}{m_1} = -\frac{1}{3} \)
Therefore, the equation of the normal at (1, -5) is
y - (-5) = \( -\frac{1}{3} \)(x - 1)
Therefore, 3y + 15 = -x + 1
Therefore, x + 3y + 14 = 0
Hence, the equations of tangent and normal are 3x - y - 8 = 0 and x + 3y + 14 = 0 respectively.
In simple words: To solve this, equate the derivative of the curve (slope of the tangent) to the slope of the given parallel line to find the point of tangency. Then use this point and the calculated slopes to construct the equations for both the tangent and the normal.

🎯 Exam Tip: A crucial step in such problems is correctly finding the point of tangency by using the condition of parallelism. Double-check your differentiation and algebraic solutions for \( x_1 \) and \( y_1 \) to avoid downstream errors in the final equations.