Tangents and Normal: Common Parametric Coordinates on a Curve & Diagrams

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Tangents and Normals is part of Unit 3 Calculus, under Chapter 6 Applications of Derivatives. Calculus is one of the most important sections in this year’s Class 12 CBSE Boards with a total weightage of 35 marks out of 80. This means that over 40% of this year's questions will be derived from Calculus. The chapters included here are: Continuity and Differentiability, Applications of Derivatives, Integrals, Applications of the Integrals, Differential Integrations

Table of Content

Introduction to Tangents and Normals
Common Parametric Coordinates on a Curve
Diagram Explaining Tangents and Normal
Things to Remember
Sample Questions

Key Terms: Intervals, Derivatives, function, Maxima and Minima, Applications of Derivatives, Tangents and normal.

Read Also: Maxima and Minima

Introduction to Tangents and Normals

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A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
Given a function y = f(x), the equation of the tangent for this curve at x = x₀ can be derived in a below-mentioned way:
Find out the derivative of the curve at the point x = x₀ : To find this one needs to dxdy⌋x=x₀. Assume this value m, in analogy to the slope of a straight line.
Find the equation of the straight line getting through the point (x₀, y(x₀)) with slope m. This is direct and can be derived as

Also Read: Differential Equations

y–y₀= m( x–x₀)

Tangent and Normal

Slope of tangent (at x=x0) m=dy/dx||_x=x0

A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got, m×n = -1
Steps for finding the normal to a given curve y = f(x) at a point _{x = x0}:
Find out the derivative of the curve at the point x = x₀:
This first step is the same as we did for finding the equation of the tangent to the curve that is m = dy/dx⌋x=x₀
The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m

The video below explains this:

Derivatives of Function in Parametric Form Detailed Video Explanation:

Read More: Approximations

Common Parametric Coordinates on a Curve

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For x^2/3 + y^2/3 = a^2/3, assume the parametric coordinates x = a cos³θ and y = a sin³θ.
For √x + √y = √a, assume x = cos⁴θ and y = a sin⁴θ
For xⁿ/aⁿ + yⁿ/bⁿ = 1, assume x = a (sin θ)^2/n and y = b(sin θ)^2/n.
For y² = x³, take x = t² and y = t³.

Also Read:

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Diagram Explaining Tangents and Normal

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Diagram Explaining Tangents and Normal

Diagram Explaining Tangents and Normal

PT is the tangent to the curve y = f(x) at the point P(x₁, y₁).
PN is normal to the curve at P.
The slope of the tangent at P(x₁, y₁) is, [dy/dx]_(x1,y1).
The slope of the normal at P(x₁, y₁) is, -1/[dy/dx]_(x1,y1).

Hence the equation of the tangent PT is, y - y₁ = [dy/dx]_(x1,y1) (x - x₁),

And the equation of the normal PN is, y - y₁ = -1/[dy/dx]_(x1,y1) (x - x₁)

If (dy/dx)_(x1,y1) = 0, then the equation of the normal would be x = x₁.

If the equation of the curve is in the parametric form x = f(t) and y = g(t), then

dy/dx = dy/dt/dx/dt = (g' (t))/(f' (t)).

Thus the equations of the tangent and the normal are

y - g(t) = (g' (t))/(f' (t)) (x - f(t))

And

f'(t)[x - f(t)] + g'(t)[y - g(t)] = 0.

Also Read: Differentiation and Integration Formula

Things to Remember

If dy/dx at (x₁, y₁) = 0 then the tangent is parallel to x-axis.
If the tangent is parallel to ax + by + c = 0 then dy/dx = -a/b
If dy/dx at (x₁, y₁) → ∞ i.e. dx/dy at (x₁, y₁) = 0 then the tangent is perpendicular to the x- axis.
If the tangent at P(x₁, y₁) is equally inclined to the coordinate axis, then dy/dx at (x₁, y₁) = ± 1.
If the tangent makes equal non-zero intercepts on the coordinate axis, then dy/dx at (x₁, y₁) = -1.
If the tangent cuts off from the coordinate axis equal to the distance from the origin then dy/dx = ±1.

Read More: logarithmic functions

Sample Questions

Ques. What are the four kinds of slopes? (1 mark)

Ans: The four different kinds of slopes are zero, undefined, positive, and negative.

Ques. Is gradient the same as a slope? (1 mark)

Ans: Gradient is the degree of a graph’s steepness at any point. Slope refers to the graph’s gradient at any point. Hence both are the same.

Ques. What is the difference between tangent and normal? (1 mark)

Ans: A tangent is a straight line whose extension begins from a point on a curve, with a derivative equal to the curve’s derivative existing at that particular point. A normal is a straight line whose extension begins from a curve’s point such that it is perpendicular to the point’s tangent.

Ques. How to find the tangent? (4 mark)

Ans: To find the tangent, you need to -

Draw the tangent line.
Take the first derivative, to find the equation for the tangent line’s slope.
Enter the x value belonging to the point.
In point-slope, write the equation of the tangent line.
Lastly, confirm the equation on the graph.

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Tangents and Normal: Common Parametric Coordinates on a Curve & Diagrams

Introduction to Tangents and Normals

Derivatives of Function in Parametric Form Detailed Video Explanation:

Common Parametric Coordinates on a Curve

Diagram Explaining Tangents and Normal

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

2.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

4.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

5.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

6.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

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Tangents and Normal: Common Parametric Coordinates on a Curve & Diagrams

Introduction to Tangents and Normals

Derivatives of Function in Parametric Form Detailed Video Explanation:

Common Parametric Coordinates on a Curve

Diagram Explaining Tangents and Normal

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1. Find: If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

2.If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3.Evaluate:\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

4.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

5.Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

6. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

2.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

4.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

5.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

6.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]