Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 7.2

Question 1. Find the slope of the tangent to the following curves at the respective given points.
(i) \( y = x^4 + 2x^2 - x \) at \( x = 1 \)
(ii) \( x = a \cos^3 t, y = b \sin^3 t \) at \( t = \frac { \pi }{ 2 } \)
Answer:
(i) We are given the curve \( y = x^4 + 2x^2 - x \).
First, differentiate \( y \) with respect to \( x \) to find the slope of the tangent:
\( \frac { dy }{ dx } = 4x^3 + 4x - 1 \)
Now, substitute the given point \( x = 1 \) into the derivative to find the slope at that point:
\( \left( \frac { dy }{ dx } \right)_{(x=1)} = 4(1)^3 + 4(1) - 1 \)
\( = 4 + 4 - 1 \)
\( = 7 \)
So, the slope of the tangent for curve (i) at \( x = 1 \) is 7.

(ii) We are given parametric equations for the curve: \( x = a \cos^3 t \) and \( y = b \sin^3 t \).
We need to differentiate both \( x \) and \( y \) with respect to \( t \).
\( \frac { dx }{ dt } = a \cdot 3 \cos^2 t \cdot (-\sin t) = -3a \cos^2 t \sin t \)
\( \frac { dy }{ dt } = b \cdot 3 \sin^2 t \cdot (\cos t) = 3b \sin^2 t \cos t \)
Next, find \( \frac { dy }{ dx } \) using the chain rule:
\( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } = \frac { 3b \sin^2 t \cos t }{ -3a \cos^2 t \sin t } \)
Simplify the expression:
\( \frac { dy }{ dx } = -\frac { b \sin t }{ a \cos t } = -\frac { b }{ a } \tan t \)
Now, substitute the given value \( t = \frac { \pi }{ 2 } \) to find the slope of the tangent at that point:
\( \left( \frac { dy }{ dx } \right)_{(t=\pi/2)} = -\frac { b }{ a } \tan \frac { \pi }{ 2 } \)
Since \( \tan \frac { \pi }{ 2 } \) is undefined (or approaches infinity), the slope of the tangent is infinite.
The tangent line is vertical when its slope is infinite.
In simple words: For part (i), we find the derivative of the curve and put \( x=1 \) into it, which gives us the slope of the tangent line. For part (ii), we use a special rule for curves defined by 't' to find the slope. When we put in \( t = \pi/2 \), the slope becomes infinitely large, meaning the tangent line stands straight up.

🎯 Exam Tip: Remember that an undefined slope means the tangent line is vertical, and a zero slope means the tangent line is horizontal. Always be careful with trigonometric values at special angles like \( \pi/2 \).

Question 2. Find the point on the curve \( y = x^2 - 5x + 4 \) at which the tangent is parallel to the line \( 3x + y = 7 \).
Answer:
The given curve is \( y = x^2 - 5x + 4 \).
First, differentiate the curve with respect to \( x \) to find the slope of the tangent:
\( \frac { dy }{ dx } = 2x - 5 \)
The given line is \( 3x + y = 7 \). To find its slope, we can rewrite it in the form \( y = mx + c \):
\( y = -3x + 7 \)
The slope of this line is \( m = -3 \).
Since the tangent to the curve is parallel to this line, their slopes must be equal.
So, we set the slope of the tangent equal to the slope of the line:
\( 2x - 5 = -3 \)
Now, solve for \( x \):
\( 2x = -3 + 5 \)
\( 2x = 2 \)
\( x = 1 \)
To find the corresponding \( y \)-coordinate, substitute \( x = 1 \) back into the original curve equation:
\( y = (1)^2 - 5(1) + 4 \)
\( y = 1 - 5 + 4 \)
\( y = 0 \)
Therefore, the point on the curve where the tangent is parallel to the given line is \( (1, 0) \). This point is where the tangent has the same direction as the given line.
In simple words: We find how steeply the curve is rising at any point using differentiation. We also find how steeply the given line is rising. Since the tangent and the line are parallel, their steepness (slopes) must be the same. We use this to find the 'x' value, then put it back into the curve equation to find the 'y' value.

🎯 Exam Tip: Remember that parallel lines have equal slopes. Always find the derivative of the curve (slope of tangent) and the slope of the given line, then equate them.

Question 3. Find the points on curve \( y = x^3 - 6x^2 + x + 3 \) where the normal is parallel to the line \( x + y = 1729 \).
Answer:
The given curve is \( y = x^3 - 6x^2 + x + 3 \).
First, find the slope of the tangent by differentiating \( y \) with respect to \( x \):
\( \frac { dy }{ dx } = 3x^2 - 12x + 1 \)
The slope of the normal to the curve is the negative reciprocal of the slope of the tangent:
Slope of normal \( = -\frac { 1 }{ \frac { dy }{ dx } } = -\frac { 1 }{ 3x^2 - 12x + 1 } \)
The given line is \( x + y = 1729 \). Rewrite it to find its slope:
\( y = -x + 1729 \)
The slope of this line is \( -1 \).
Since the normal to the curve is parallel to this line, their slopes must be equal.
So, we set the slope of the normal equal to \( -1 \):
\( -\frac { 1 }{ 3x^2 - 12x + 1 } = -1 \)
This means:
\( \frac { 1 }{ 3x^2 - 12x + 1 } = 1 \)
\( 3x^2 - 12x + 1 = 1 \)
\( 3x^2 - 12x = 0 \)
Factor out \( 3x \):
\( 3x(x - 4) = 0 \)
This gives two possible values for \( x \):
\( 3x = 0 \implies x = 0 \)
\( x - 4 = 0 \implies x = 4 \)
Now, find the corresponding \( y \)-coordinates for each \( x \) value using the original curve equation:
When \( x = 0 \):
\( y = (0)^3 - 6(0)^2 + 0 + 3 = 3 \)
So, the first point is \( (0, 3) \).
When \( x = 4 \):
\( y = (4)^3 - 6(4)^2 + 4 + 3 \)
\( y = 64 - 6(16) + 4 + 3 \)
\( y = 64 - 96 + 4 + 3 \)
\( y = -25 \)
So, the second point is \( (4, -25) \).
Thus, the points on the curve where the normal is parallel to the line \( x + y = 1729 \) are \( (0, 3) \) and \( (4, -25) \). The normal is a line perpendicular to the tangent at that point.
In simple words: First, we find the formula for the steepness of the tangent line on the curve. Then, we find the steepness of the line that is perpendicular to the tangent (called the normal). We are told this normal line is parallel to another given line, so their steepness must be the same. We use this to find the 'x' values, and then plug them back into the curve's equation to get the 'y' values, giving us the points.

🎯 Exam Tip: Remember that the slope of the normal is the negative reciprocal of the slope of the tangent. If two lines are parallel, their slopes are equal.

Question 4. Find the points on the curve \( y^2 - 4xy = x^2 + 5 \) for which the tangent is horizontal.
Answer:
The given curve is \( y^2 - 4xy = x^2 + 5 \) (Equation 1).
For the tangent to be horizontal, its slope \( \frac { dy }{ dx } \) must be zero.
We need to differentiate the curve implicitly with respect to \( x \):
\( \frac { d }{ dx } (y^2) - \frac { d }{ dx } (4xy) = \frac { d }{ dx } (x^2) + \frac { d }{ dx } (5) \)
\( 2y \frac { dy }{ dx } - 4 \left( x \frac { dy }{ dx } + y \cdot 1 \right) = 2x + 0 \)
\( 2y \frac { dy }{ dx } - 4x \frac { dy }{ dx } - 4y = 2x \)
Group terms with \( \frac { dy }{ dx } \):
\( (2y - 4x) \frac { dy }{ dx } = 2x + 4y \)
Solve for \( \frac { dy }{ dx } \):
\( \frac { dy }{ dx } = \frac { 2x + 4y }{ 2y - 4x } = \frac { 2(x + 2y) }{ 2(y - 2x) } = \frac { x + 2y }{ y - 2x } \)
Now, set the slope of the tangent to zero for a horizontal tangent:
\( \frac { x + 2y }{ y - 2x } = 0 \)
This means the numerator must be zero (as long as the denominator is not also zero):
\( x + 2y = 0 \)
From this, we can express \( x \) in terms of \( y \):
\( x = -2y \)
Substitute this expression for \( x \) back into the original curve equation (Equation 1):
\( y^2 - 4(-2y)y = (-2y)^2 + 5 \)
\( y^2 + 8y^2 = 4y^2 + 5 \)
\( 9y^2 = 4y^2 + 5 \)
\( 9y^2 - 4y^2 = 5 \)
\( 5y^2 = 5 \)
\( y^2 = 1 \)
Taking the square root, we get two possible values for \( y \):
\( y = \pm 1 \)
Now, find the corresponding \( x \) values using \( x = -2y \):
When \( y = 1 \):
\( x = -2(1) = -2 \)
When \( y = -1 \):
\( x = -2(-1) = 2 \)
Therefore, the points on the curve where the tangent is horizontal are \( (-2, 1) \) and \( (2, -1) \). At these points, the curve levels out for an instant.
In simple words: A horizontal tangent means the slope is zero. We find the slope of the curve using implicit differentiation. We then set this slope to zero to find a relationship between 'x' and 'y'. Plugging this relationship back into the original curve equation helps us find the exact 'x' and 'y' points where the tangent is flat.

🎯 Exam Tip: For horizontal tangents, remember to set \( \frac { dy }{ dx } = 0 \). For vertical tangents, set the denominator of \( \frac { dy }{ dx } \) to zero (if \( \frac { dy }{ dx } \) is a fraction) and ensure the numerator is not also zero.

Question 5. Find the tangent and normal to the following curves at the given points on the curve.
(i) \( y = x^2 - x^4 \) at \( (1, 0) \)
(ii) \( y = x^4 + 2e^x \) at \( (0, 2) \)
(iii) \( y = x \sin x \) at \( \left( \frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \)
(iv) \( x = \cos t, y = 2 \sin^2 t \) at \( t = \frac { \pi }{ 3 } \)
Answer:
(i) Given curve: \( y = x^2 - x^4 \) at point \( (1, 0) \).
First, find the slope of the tangent by differentiating \( y \) with respect to \( x \):
\( \frac { dy }{ dx } = 2x - 4x^3 \)
Now, calculate the slope 'm' at the given point \( (1, 0) \):
\( m = \left( \frac { dy }{ dx } \right)_{(1, 0)} = 2(1) - 4(1)^3 = 2 - 4 = -2 \)
The slope of the normal is the negative reciprocal of the tangent's slope:
Slope of normal \( = -\frac { 1 }{ m } = -\frac { 1 }{ -2 } = \frac { 1 }{ 2 } \)
Equation of the tangent using \( y - y_1 = m(x - x_1) \):
\( y - 0 = -2(x - 1) \)
\( y = -2x + 2 \)
\( 2x + y - 2 = 0 \)
Equation of the normal using \( y - y_1 = -\frac { 1 }{ m }(x - x_1) \):
\( y - 0 = \frac { 1 }{ 2 }(x - 1) \)
\( 2y = x - 1 \)
\( x - 2y - 1 = 0 \)

(ii) Given curve: \( y = x^4 + 2e^x \) at point \( (0, 2) \).
First, find the slope of the tangent by differentiating \( y \) with respect to \( x \):
\( \frac { dy }{ dx } = 4x^3 + 2e^x \)
Now, calculate the slope 'm' at the given point \( (0, 2) \):
\( m = \left( \frac { dy }{ dx } \right)_{(0, 2)} = 4(0)^3 + 2e^0 = 0 + 2(1) = 2 \)
The slope of the normal is the negative reciprocal of the tangent's slope:
Slope of normal \( = -\frac { 1 }{ m } = -\frac { 1 }{ 2 } \)
Equation of the tangent using \( y - y_1 = m(x - x_1) \):
\( y - 2 = 2(x - 0) \)
\( y - 2 = 2x \)
\( 2x - y + 2 = 0 \)
Equation of the normal using \( y - y_1 = -\frac { 1 }{ m }(x - x_1) \):
\( y - 2 = -\frac { 1 }{ 2 }(x - 0) \)
\( 2(y - 2) = -x \)
\( 2y - 4 = -x \)
\( x + 2y - 4 = 0 \)

(iii) Given curve: \( y = x \sin x \) at point \( \left( \frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \).
First, find the slope of the tangent by differentiating \( y \) with respect to \( x \) using the product rule:
\( \frac { dy }{ dx } = x \frac { d }{ dx } (\sin x) + \sin x \frac { d }{ dx } (x) \)
\( \frac { dy }{ dx } = x \cos x + \sin x \)
Now, calculate the slope 'm' at the given point \( \left( \frac { \pi }{ 2 }, \frac { \pi }{ 2 } \right) \):
\( m = \left( \frac { dy }{ dx } \right)_{(\pi/2, \pi/2)} = \frac { \pi }{ 2 } \cos \left( \frac { \pi }{ 2 } \right) + \sin \left( \frac { \pi }{ 2 } \right) \)
\( = \frac { \pi }{ 2 } (0) + 1 \)
\( = 1 \)
The slope of the normal is the negative reciprocal of the tangent's slope:
Slope of normal \( = -\frac { 1 }{ m } = -\frac { 1 }{ 1 } = -1 \)
Equation of the tangent using \( y - y_1 = m(x - x_1) \):
\( y - \frac { \pi }{ 2 } = 1 \left( x - \frac { \pi }{ 2 } \right) \)
\( y - \frac { \pi }{ 2 } = x - \frac { \pi }{ 2 } \)
\( x - y = 0 \)
Equation of the normal using \( y - y_1 = -\frac { 1 }{ m }(x - x_1) \):
\( y - \frac { \pi }{ 2 } = -1 \left( x - \frac { \pi }{ 2 } \right) \)
\( y - \frac { \pi }{ 2 } = -x + \frac { \pi }{ 2 } \)
\( x + y - \pi = 0 \)

(iv) Given parametric curve: \( x = \cos t, y = 2 \sin^2 t \) at \( t = \frac { \pi }{ 3 } \).
First, find the coordinates of the point at \( t = \frac { \pi }{ 3 } \):
\( x = \cos \left( \frac { \pi }{ 3 } \right) = \frac { 1 }{ 2 } \)
\( y = 2 \sin^2 \left( \frac { \pi }{ 3 } \right) = 2 \left( \frac { \sqrt{3} }{ 2 } \right)^2 = 2 \left( \frac { 3 }{ 4 } \right) = \frac { 3 }{ 2 } \)
So, the point is \( \left( \frac { 1 }{ 2 }, \frac { 3 }{ 2 } \right) \).
Next, find \( \frac { dx }{ dt } \) and \( \frac { dy }{ dt } \):
\( \frac { dx }{ dt } = -\sin t \)
\( \frac { dy }{ dt } = 2 \cdot 2 \sin t \cos t = 4 \sin t \cos t \)
Now, find the slope of the tangent \( m = \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \):
\( m = \frac { 4 \sin t \cos t }{ -\sin t } = -4 \cos t \)
Calculate the slope 'm' at \( t = \frac { \pi }{ 3 } \):
\( m = -4 \cos \left( \frac { \pi }{ 3 } \right) = -4 \left( \frac { 1 }{ 2 } \right) = -2 \)
The slope of the normal is the negative reciprocal of the tangent's slope:
Slope of normal \( = -\frac { 1 }{ m } = -\frac { 1 }{ -2 } = \frac { 1 }{ 2 } \)
Equation of the tangent using \( y - y_1 = m(x - x_1) \):
\( y - \frac { 3 }{ 2 } = -2 \left( x - \frac { 1 }{ 2 } \right) \)
Multiply by 2 to clear fractions:
\( 2y - 3 = -4 \left( x - \frac { 1 }{ 2 } \right) \)
\( 2y - 3 = -4x + 2 \)
\( 4x + 2y - 5 = 0 \)
Equation of the normal using \( y - y_1 = -\frac { 1 }{ m }(x - x_1) \):
\( y - \frac { 3 }{ 2 } = \frac { 1 }{ 2 } \left( x - \frac { 1 }{ 2 } \right) \)
Multiply by 2 to clear fractions:
\( 2 \left( y - \frac { 3 }{ 2 } \right) = x - \frac { 1 }{ 2 } \)
\( 2y - 3 = x - \frac { 1 }{ 2 } \)
Multiply by 2 again:
\( 4y - 6 = 2x - 1 \)
\( 2x - 4y + 5 = 0 \)
In simple words: For each part, we first find the slope of the curve (the tangent) at the given point by differentiating. Then, we use the point and the slope to write the equation of the tangent line. For the normal line, we take the negative inverse of the tangent's slope and use the same point to write its equation. This helps us describe how the curve behaves at specific spots.

🎯 Exam Tip: Always state the formula for the tangent and normal equations. Double-check your differentiation and algebraic simplification steps, especially when dealing with negative reciprocals and fractions.

Question 6. Find the equations of the tangents to the curve \( y = 1 + x^3 \) for which the tangent is orthogonal with the line \( x + 12y = 12 \).
Answer:
The given curve is \( y = 1 + x^3 \).
First, find the slope of the tangent by differentiating \( y \) with respect to \( x \):
\( \frac { dy }{ dx } = 3x^2 \)
The given line is \( x + 12y = 12 \). To find its slope, rewrite it in \( y = mx + c \) form:
\( 12y = -x + 12 \)
\( y = -\frac { 1 }{ 12 } x + 1 \)
The slope of this line is \( -\frac { 1 }{ 12 } \).
Since the tangent to the curve is orthogonal (perpendicular) to this line, the product of their slopes must be \( -1 \).
Slope of tangent \( \times \) Slope of line \( = -1 \)
Slope of tangent \( = -\frac { 1 }{ (-\frac { 1 }{ 12 }) } = 12 \)
Now, set the slope of the tangent to the curve equal to 12:
\( 3x^2 = 12 \)
\( x^2 = \frac { 12 }{ 3 } \)
\( x^2 = 4 \)
Taking the square root, we get two possible values for \( x \):
\( x = \pm 2 \)
Now, find the corresponding \( y \)-coordinates for each \( x \) value using the original curve equation \( y = 1 + x^3 \):
When \( x = 2 \):
\( y = 1 + (2)^3 = 1 + 8 = 9 \)
So, the first point is \( (2, 9) \).
When \( x = -2 \):
\( y = 1 + (-2)^3 = 1 - 8 = -7 \)
So, the second point is \( (-2, -7) \).
Now, we find the equations of the tangents at these two points with a slope of 12.
Equation of tangent at \( (2, 9) \) with \( m = 12 \):
\( y - y_1 = m(x - x_1) \)
\( y - 9 = 12(x - 2) \)
\( y - 9 = 12x - 24 \)
\( 12x - y - 15 = 0 \)
Equation of tangent at \( (-2, -7) \) with \( m = 12 \):
\( y - (-7) = 12(x - (-2)) \)
\( y + 7 = 12(x + 2) \)
\( y + 7 = 12x + 24 \)
\( 12x - y + 17 = 0 \)
Therefore, the equations of the two tangents are \( 12x - y - 15 = 0 \) and \( 12x - y + 17 = 0 \). The orthogonal condition helps find the specific slope of the tangent.
In simple words: First, we figure out how steep the tangent line should be by using the information that it's perpendicular to another given line. Then, we find where on our curve the tangent has this specific steepness. Once we have these points and the slope, we can write the equation for each tangent line.

🎯 Exam Tip: For orthogonal lines, the product of their slopes is \( -1 \). Make sure to find all possible \( x \) values from the equation of the slope, as there can be multiple tangent lines.

Question 7. Find the equations of the tangents to the curve \( y = -\frac { x+1 }{ x-1 } \) which are parallel to the line \( x + 2y = 6 \).
Answer:
The given curve is \( y = -\frac { x+1 }{ x-1 } \).
First, find the slope of the tangent \( \frac { dy }{ dx } \) using the quotient rule \( \left( \frac { u }{ v } \right)' = \frac { u'v - uv' }{ v^2 } \):
Let \( u = -(x+1) = -x-1 \) and \( v = x-1 \). Then \( u' = -1 \) and \( v' = 1 \).
\( \frac { dy }{ dx } = \frac { (-1)(x-1) - (-x-1)(1) }{ (x-1)^2 } \)
\( \frac { dy }{ dx } = \frac { -x+1 + x+1 }{ (x-1)^2 } \)
\( \frac { dy }{ dx } = \frac { 2 }{ (x-1)^2 } \)
The given line is \( x + 2y = 6 \). To find its slope, rewrite it in the form \( y = mx + c \):
\( 2y = -x + 6 \)
\( y = -\frac { 1 }{ 2 } x + 3 \)
The slope of this line is \( -\frac { 1 }{ 2 } \).
Since the tangent to the curve is parallel to this line, their slopes must be equal.
So, we set the slope of the tangent equal to \( -\frac { 1 }{ 2 } \):
\( \frac { 2 }{ (x-1)^2 } = -\frac { 1 }{ 2 } \)
Cross-multiply:
\( 2 \cdot 2 = -1 \cdot (x-1)^2 \)
\( 4 = -(x-1)^2 \)
\( (x-1)^2 = -4 \)
Since the square of a real number cannot be negative, there is no real value of \( x \) that satisfies this equation. This means there are no points on the curve where the tangent is parallel to the given line.
Therefore, there are no tangents to the curve that are parallel to the line \( x + 2y = 6 \). This indicates that the slope \( \frac{2}{(x-1)^2} \) is always positive, and can never be \( -1/2 \).
In simple words: We find the slope of the curve using differentiation and the slope of the given line. Since the tangent needs to be parallel to the line, their slopes must be the same. When we try to find the 'x' values where this happens, we end up with a math problem that has no real solution. This tells us there are no such tangent lines.

🎯 Exam Tip: Always check if your solution for \( x \) yields real numbers. If you encounter a situation like \( (x-a)^2 = \text{negative number} \), it means no real solutions exist, and therefore no such points or tangents.

Question 8. Find the equation of tangent and normal to the curve given by \( x = 7 \cos t \) and \( y = 2 \sin t, t \in \mathbb{R} \) at any point on the curve.
Answer:
The given parametric curve is \( x = 7 \cos t \) and \( y = 2 \sin t \).
First, find \( \frac { dx }{ dt } \) and \( \frac { dy }{ dt } \):
\( \frac { dx }{ dt } = \frac { d }{ dt } (7 \cos t) = -7 \sin t \)
\( \frac { dy }{ dt } = \frac { d }{ dt } (2 \sin t) = 2 \cos t \)
The slope of the tangent 'm' at any point \( t \) is given by \( \frac { dy }{ dx } = \frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } \):
\( m = \frac { 2 \cos t }{ -7 \sin t } = -\frac { 2 }{ 7 } \cot t \)
The coordinates of any point on the curve are \( (x_1, y_1) = (7 \cos t, 2 \sin t) \).
Equation of the tangent at \( (7 \cos t, 2 \sin t) \) with slope \( m = -\frac { 2 }{ 7 } \cot t \):
\( y - y_1 = m(x - x_1) \)
\( y - 2 \sin t = -\frac { 2 }{ 7 } \cot t (x - 7 \cos t) \)
Replace \( \cot t \) with \( \frac { \cos t }{ \sin t } \):
\( y - 2 \sin t = -\frac { 2 \cos t }{ 7 \sin t } (x - 7 \cos t) \)
Multiply both sides by \( 7 \sin t \) to clear the denominator:
\( 7 \sin t (y - 2 \sin t) = -2 \cos t (x - 7 \cos t) \)
\( 7y \sin t - 14 \sin^2 t = -2x \cos t + 14 \cos^2 t \)
Rearrange the terms to form the equation of the tangent:
\( 2x \cos t + 7y \sin t - 14 \sin^2 t - 14 \cos^2 t = 0 \)
\( 2x \cos t + 7y \sin t - 14 (\sin^2 t + \cos^2 t) = 0 \)
Since \( \sin^2 t + \cos^2 t = 1 \):
\( 2x \cos t + 7y \sin t - 14(1) = 0 \)
\( 2x \cos t + 7y \sin t - 14 = 0 \)
Now, find the slope of the normal, which is the negative reciprocal of the tangent's slope:
Slope of normal \( = -\frac { 1 }{ m } = -\frac { 1 }{ -\frac { 2 }{ 7 } \cot t } = \frac { 7 }{ 2 } \tan t \)
Equation of the normal at \( (7 \cos t, 2 \sin t) \) with slope \( \frac { 7 }{ 2 } \tan t \):
\( y - y_1 = -\frac { 1 }{ m }(x - x_1) \)
\( y - 2 \sin t = \frac { 7 }{ 2 } \tan t (x - 7 \cos t) \)
Replace \( \tan t \) with \( \frac { \sin t }{ \cos t } \):
\( y - 2 \sin t = \frac { 7 \sin t }{ 2 \cos t } (x - 7 \cos t) \)
Multiply both sides by \( 2 \cos t \):
\( 2 \cos t (y - 2 \sin t) = 7 \sin t (x - 7 \cos t) \)
\( 2y \cos t - 4 \sin t \cos t = 7x \sin t - 49 \sin t \cos t \)
Rearrange the terms:
\( 7x \sin t - 2y \cos t - 49 \sin t \cos t + 4 \sin t \cos t = 0 \)
\( 7x \sin t - 2y \cos t - 45 \sin t \cos t = 0 \)
This gives the equations for the tangent and normal lines at any point on the curve. These equations change as 't' changes, showing the dynamic nature of the curve.
In simple words: We find the slope of the curve using differentiation. Since the curve is given using 't', we find the slope using a special formula. Then, using the slope and the point, we write the equation for the tangent line. For the normal line, which is perpendicular, we use the negative inverse of the slope and write its equation.

🎯 Exam Tip: When dealing with parametric equations, remember to use \( \frac{ dy }{ dx } = \frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } } \). Keep trigonometric identities like \( \sin^2 t + \cos^2 t = 1 \) in mind for simplification.

Question 9. Find the angle between the rectangular hyperbola \( xy = 2 \) and the parabola \( x^2 + 4y = 0 \).
Answer:
To find the angle between two curves, we first need to find their points of intersection and then the slopes of their tangents at those points.
Given curves:
1. Rectangular hyperbola: \( xy = 2 \implies y = \frac { 2 }{ x } \) (Equation 1)
2. Parabola: \( x^2 + 4y = 0 \) (Equation 2)
Substitute \( y \) from Equation 1 into Equation 2 to find the intersection points:
\( x^2 + 4 \left( \frac { 2 }{ x } \right) = 0 \)
\( x^2 + \frac { 8 }{ x } = 0 \)
Multiply by \( x \) (assuming \( x \neq 0 \)):
\( x^3 + 8 = 0 \)
\( x^3 = -8 \)
\( x = -2 \)
Now, find the corresponding \( y \) value using Equation 1:
\( y = \frac { 2 }{ -2 } = -1 \)
So, the point of intersection is \( (-2, -1) \).

Next, find the slopes of the tangents for each curve at \( (-2, -1) \).
For the hyperbola \( xy = 2 \):
Differentiate implicitly with respect to \( x \):
\( x \frac { dy }{ dx } + y \cdot 1 = 0 \)
\( \frac { dy }{ dx } = -\frac { y }{ x } \)
At the point \( (-2, -1) \), the slope \( m_1 \) is:
\( m_1 = -\frac { (-1) }{ (-2) } = -\frac { 1 }{ 2 } \)

For the parabola \( x^2 + 4y = 0 \):
Differentiate implicitly with respect to \( x \):
\( 2x + 4 \frac { dy }{ dx } = 0 \)
\( 4 \frac { dy }{ dx } = -2x \)
\( \frac { dy }{ dx } = -\frac { 2x }{ 4 } = -\frac { x }{ 2 } \)
At the point \( (-2, -1) \), the slope \( m_2 \) is:
\( m_2 = -\frac { (-2) }{ 2 } = 1 \)

Finally, find the angle \( \theta \) between the two curves using the formula:
\( \tan \theta = \left| \frac { m_1 - m_2 }{ 1 + m_1 m_2 } \right| \)
\( \tan \theta = \left| \frac { -\frac { 1 }{ 2 } - 1 }{ 1 + \left( -\frac { 1 }{ 2 } \right) (1) } \right| \)
\( \tan \theta = \left| \frac { -\frac { 1 }{ 2 } - \frac { 2 }{ 2 } }{ 1 - \frac { 1 }{ 2 } } \right| \)
\( \tan \theta = \left| \frac { -\frac { 3 }{ 2 } }{ \frac { 1 }{ 2 } } \right| \)
\( \tan \theta = \left| -\frac { 3 }{ 2 } \cdot \frac { 2 }{ 1 } \right| \)
\( \tan \theta = |-3| = 3 \)
Therefore, the angle between the curves is \( \theta = \tan^{-1} (3) \). This angle describes how steeply the curves cross each other.
In simple words: First, we find the point where the two curves meet. Then, at that meeting point, we find how steep each curve is (their slopes). Finally, we use a special formula that takes these two slopes to calculate the angle at which the curves cross each other.

🎯 Exam Tip: To find the angle between curves, always remember these steps: 1) Find intersection points. 2) Find \( \frac{dy}{dx} \) for each curve. 3) Calculate slopes \( m_1 \) and \( m_2 \) at the intersection points. 4) Use the angle formula \( \tan \theta = \left| \frac { m_1 - m_2 }{ 1 + m_1 m_2 } \right| \).

Question 10. Show that the two curves \( x^2 - y^2 = r^2 \) and \( xy = c^2 \) where \( c, r \) are constants, cut orthogonally.
Answer:
Two curves cut orthogonally if their tangents at the point of intersection are perpendicular. This means the product of their slopes at the intersection point must be \( -1 \).
Given curves:
1. \( x^2 - y^2 = r^2 \) (Equation 1)
2. \( xy = c^2 \) (Equation 2)

First, find the slope of the tangent for the first curve by differentiating Equation 1 implicitly with respect to \( x \):
\( \frac { d }{ dx } (x^2) - \frac { d }{ dx } (y^2) = \frac { d }{ dx } (r^2) \)
\( 2x - 2y \frac { dy }{ dx } = 0 \)
\( 2x = 2y \frac { dy }{ dx } \)
\( \frac { dy }{ dx } = \frac { 2x }{ 2y } = \frac { x }{ y } \)
Let \( m_1 \) be the slope of the tangent for the first curve at an intersection point \( (x_0, y_0) \). Then \( m_1 = \frac { x_0 }{ y_0 } \).

Next, find the slope of the tangent for the second curve by differentiating Equation 2 implicitly with respect to \( x \):
\( \frac { d }{ dx } (xy) = \frac { d }{ dx } (c^2) \)
Using the product rule on \( xy \):
\( x \frac { dy }{ dx } + y \cdot 1 = 0 \)
\( x \frac { dy }{ dx } = -y \)
\( \frac { dy }{ dx } = -\frac { y }{ x } \)
Let \( m_2 \) be the slope of the tangent for the second curve at the same intersection point \( (x_0, y_0) \). Then \( m_2 = -\frac { y_0 }{ x_0 } \).

Now, we check the product of the slopes \( m_1 m_2 \) at the intersection point \( (x_0, y_0) \):
\( m_1 m_2 = \left( \frac { x_0 }{ y_0 } \right) \left( -\frac { y_0 }{ x_0 } \right) \)
\( m_1 m_2 = -1 \)
Since the product of the slopes of the tangents at their intersection point is \( -1 \), the two curves cut orthogonally. This condition proves that the lines are perpendicular, showcasing a fundamental property of these curve types.
In simple words: To show curves cross at a right angle (orthogonally), we need to show that their tangent lines are perpendicular at the point where they meet. We find the slope of the tangent for each curve using differentiation. Then, we multiply these slopes together. If the result is \( -1 \), it means the tangent lines are perpendicular, and thus the curves cross orthogonally.

🎯 Exam Tip: The key to proving orthogonal intersection is to show that \( m_1 m_2 = -1 \) at any point of intersection. Remember to use implicit differentiation correctly for both curve equations.