Tangents and Normals – Application of Derivatives

1.0Introduction to Tangents and Normals as an Application of Derivatives

In calculus, the concept of a derivative is far more than just a formula for differentiation. Geometrically, the derivative of a function at a point represents the instantaneous rate of change, which translates to the slope of the tangent line to the curve at that specific point. This fundamental geometric interpretation is the basis for a crucial section of calculus: the study of tangents and normals.

A tangent line is a straight line that grazes a curve at a single point, called the point of tangency, sharing the same instantaneous direction as the curve at that point. A normal line, on the other hand, is a straight line that is perpendicular to the tangent line at the point of tangency.

Mastering this topic is essential for JEE aspirants as it forms the bedrock for solving a wide range of problems, from curve sketching and optimization to understanding physical concepts like velocity and acceleration.

2.0What are Tangents and Normals? 

Imagine you're walking along a curved path. At any point on your walk, the direction you are facing is a straight line that just touches the path. That straight line is the tangent.

• A Tangent is a line that touches a curve at a single point and shares the same direction as the curve at that exact moment.
• A Normal is a line that is perfectly perpendicular (at a 90-degree angle) to the tangent at the point of contact.

In calculus, we use derivatives to figure out the exact direction of the tangent at any point on a curve. This is a powerful skill and a key application of derivatives in mathematics

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3.0The Fundamental Connection: Derivative as the Slope

This is the most important concept to grasp. In simple terms:

The derivative of a function, f′(x), gives you the slope of the tangent line to the curve y=f(x) at any point.

• Slope of the Tangent $m_T$ : If you have a curve y=f(x), the slope of the tangent at a specific point $(x_1,y_1)$ is found by calculating the derivative and plugging in the x-value. 

$m_T={\frac{dy}{dx}|}_{(x_1,y_1)}=f^{\prime }(x_1)$

• Slope of the Normal $m_N$ : Because the normal is perpendicular to the tangent, their slopes have a special relationship: their product is -1. 

$m_N=-\frac{1}{m_T}=-{\frac{1}{dy/dx}|}_{(x_1,y_1)}$

This relationship, $m_T\cdot m_N=-1$ is a fundamental rule you'll use constantly.

What if the curve is given in parametric form? If x=f(t) and y=g(t), you can still find the slope using the Chain Rule: 

$m_T=\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

4.0How to Find the Equation of a Tangent Line

Finding the equation of a line requires two things: a point on the line and its slope.

The Strategy:

1. Get the Point $(x_1,y_1)$ You'll either be given this point or you'll need to find it using the curve's equation.
2. Find the Slope $m_T$ Calculate the derivative $\frac{dy}{dx}$ and then substitute the x-value of your point into the derivative to get a numerical slope.
3. Use the Point-Slope Formula: Plug your point and slope into the classic formula: $y-y_1=m_T(x-x_1)$

That's it! This three-step process is the key to solving most problems on this topic.

5.0How to Find the Equation of a Normal Line

Finding the normal's equation follows the same logic as the tangent's, but with a different slope.

The Strategy:

1. Get the Point $(x_1,y_1)$:  Same as before, this point must be on the curve.
2. Find the Slope of the Normal $m_N$: First, find the slope of the tangent $m_T=\frac{dy}{dx}$ Then, use the relationship $m_N=-\frac{1}{m_T}$ to find the normal's slope.
3. Use the Point-Slope Formula: Plug your point and the normal's slope into the formula: $y-y_1=m_N(x-x_1)$

6.0Angle of Intersection of Two Curves

The angle between two intersecting curves is defined as the angle between their respective tangent lines at the point of intersection.

To find the angle:

1. Find the point(s) of intersection of the two curves, y=f(x) and y=g(x), by solving their equations simultaneously.
2. Calculate the slopes of the tangents to each curve at the point of intersection. Let these slopes be $m_1=f^{\prime }(x_1)\text{and}{m}_2=g^{\prime }(x_1)$.
3. Use the formula for the angle between two lines: $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

• Orthogonal Curves: If the curves intersect at a right angle $(\theta =90^{\circ })$ then tanθ is undefined, which implies the denominator is zero. This gives the condition for orthogonality: $1+m_1{m}_2=0$ or $m_1{m}_2=-1$ This is a very common JEE concept.
• Tangential Curves: If the curves touch each other, they have the same tangent at the point of intersection. This means the slopes are equal: $m_1=m_2$

7.0Special Cases: Parallel and Perpendicular Tangents/Normals

This section addresses scenarios where the tangent or normal lines have a specific orientation.

• Tangent parallel to the x-axis: The slope of the tangent is zero. Set $\frac{dy}{dx}=0$ and solve for the x-coordinates.
• Tangent parallel to the y-axis: The slope of the tangent is undefined. Set $\frac{dy}{dx}=0$ or find where the denominator of $\frac{dy}{dx}$ is zero.
• Normal parallel to the x-axis: The slope of the normal is zero. This happens when the tangent is a vertical line.
• Normal parallel to the y-axis: The slope of the normal is undefined. This happens when the tangent is a horizontal line.

8.0Length of Tangent, Normal, Subtangent, and Subnormal

These are geometric lengths related to the tangent and normal lines, often asked in JEE.

For a point $P(x_1,y_1)$ on a curve and its tangent and normal intersecting the x-axis at T and N respectively, and the projection of P on the x-axis at $M(x_1,0)$:

• $\text{Length of Subtangent}(MT):\left|\frac{y_1}{m_T}\right|=\left|\frac{y_1}{dy/dx}\right|$
• $\text{Length of Subnormal}(MN):|y_1\cdot m_T|=\left|y_1\frac{dy}{dx}\right|$
• $\text{Length of Tangent}(PT):\left|y_1\sqrt{1+(1/m_T{)}^2}\right|$
• $\text{Length of Normal}(PN):|y_1\sqrt{1+m_T^2}|$

9.0JEE-Level Problem Solving Strategies

1. Identify the Goal: Is the question asking for an equation, a point, an angle, or a length?
2. Find the Derivative: The first step in almost every problem is to calculate $\frac{dy}{dx}$ . Be careful with implicit differentiation or parametric differentiation.
3. Relate Information to Slope: Connect the given information (e.g., parallel to a line, perpendicular to another line, specific point) to the slope of the tangent or normal.
4. Set up Equations: Use the point-slope form or other relevant formulas to set up the necessary equations.
5. Solve and Verify: Solve the equations for the unknown variables (like x, y, or a constant) and always double-check your calculations.

10.0Solved Examples on Tangents and Normals

Example 1:

Find the equation of the normal to the curve $x^2+y^2=25$ at the point (3,4).

Solution:

1. Point of tangency: Given as (3,4).
2. Find derivative: Differentiate the equation implicitly with respect to x: $2x+2y\frac{dy}{dx}=0⟹\frac{dy}{dx}=-\frac{2x}{2y}=-\frac{x}{y}$
3. Calculate slope of tangent: At the point (3,4), the slope of the tangent is: $m_T=-\frac{3}{4}$
4. Calculate slope of normal: The slope of the normal is $m_N=-\frac{1}{m_T}=-\frac{1}{-3/4}=\frac{4}{3}$
5. Equation of normal: Using the point-slope form:

$$\begin{gathered} y-4=\frac{4}{3}(x-3) \\ 3(y-4)=4(x-3) \\ 3y-12=4x-12 \\ 4x-3y=0 \end{gathered}$$

Example 2:

Find the points on the curve $y=x^3-x$ at which the tangent is parallel to the line y = 2x - 5

Solution:

1. Relate slopes: The tangent is parallel to the line y = 2x - 5. The slope of this line is 2. Therefore, the slope of the tangent must also be 2.
2. Find derivative: Differentiate the curve's equation: $\frac{dy}{dx}=3x^2-1$.
3. Set slopes equal: Set the derivative equal to the required slope:

$3x^2-1=2⟹3x^2=3⟹x^2=1⟹x=\pm 1$.

1. Find points:

• If x = 1, $y=(1{)}^3-1=0$. The point is (1, 0)
• If x = -1, $y=(-1{)}^3-(-1)=-1+1=0$. The point is (-1, 0)

The required points are (1,0) and (−1,0).

11.0Practice Questions on Tangents and Normals

1. Find the equation of the tangent to the curve $y^2=4ax$ at the point $(a{t}^2,2at)$.
2. Determine the equation of the normal to the parabola $y^2=12x$ at the point (3, 6).
3. Find the slope of the tangent to the curve $y=x^3-2x+1atx=2$.
4. Find the equation of the tangent to the circle $x^2+y^2=25$ at the point (3, 4).
5. If the line y = 2x + c is tangent to the parabola $y^2=8x$. Find the value of c.
6. Find the equations of both tangents drawn to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ from the external point (0, 5).
7. Write the equation of the normal to the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ at the point $(5,\frac{8}{3})$.