Tangents and Normals: Have you ever sat on a merry-go-round? If yes, then you would understand from your experience when I tell you that the force you experience is towards the centre of the merry-go-round but your velocity (the tendency of motion) is in the way towards which your body is pointing.
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Another way of saying the same thing would be to let you know that your velocity at any point is tangential while the force at any point is normal to the circle along which you are moving. Can you draw a connection between both the ways of saying the same thing?
Don’t worry if you can’t because that’s what this branch of application of derivatives is concerned with: Finding tangents and normals to a given curve. It is a branch of great significance in finding the different maxima and minima of a function, analyzing the directions of velocity and acceleration of a moving object, finding the angles and the shortest distance between two curves and much more. Let’s jump straight into it!
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Tangent
A tangent at a point on the curve is a straight line that touches the curve at that point and whose slope is equal to the gradient/derivative of the curve at that point. From the definition, you can deduce how to find the equation of the tangent to the curve at any point. Given a function y = f(x), the equation of the tangent to this curve at x = x0 can be found in the following way:
- Find out the gradient/derivative of the curve at the point x = x0 : To do this one needs to calculate \( \frac{dy}{dx} \rfloor_{x = x_0} \). Let us call this value m, in analogy to the slope of a straight line.
- Find the equation of the straight line passing through the point (x0, y(x0)) with slope m. This is quite straightforward and can be found out as $$ \frac{y – y_1}{x – x_1} { = m} $$ You have found out the equation of the tangent to the curve at the given point!
Normal
A normal at a point on the curve is a straight line that intersects the curve at that point and is perpendicular to the tangent at that point. If its slope is given by n, and the slope of the tangent at that point or the value of the gradient/derivative at that point is given by m; then we have m×n = -1. Steps for finding the normal to a given curve y = f(x) at a point x = x0:
- Find out the gradient/derivative of the curve at the point x = x0: This first step is exactly the same as in the method of finding the equation of the tangent to the curve i.e. m = \( \frac{dy}{dx} \rfloor_{x = x_0} \)
- Find the slope ‘n’ of the normal:Â As the normal is perpendicular to the tangent, we have: $$ {n = } \frac{-1}{m} $$
- Now, find the equation of the straight line passing through the point (x0, y(x0)) with slope n. The equation is given by: $$ \frac{y – y_1}{x – x_1} { = n} $$
It might be quite noticeable that both the tangents and normals to a curve go hand in hand. Both are easily derivable from one another. Now take a look at the diagram below to visualize them better, and then proceed towards the solved example to clear your doubts.
Solved Examples on Tangents and Normals
Question 1: Consider the curve given by y = f(x) = x3 – x + 3.
- Find the equation of the line tangent to the curve at the point (1,3)
- Find the line normal to the curve at the point (1,3)
Answer : a) We can see that the point (1,3) satisfies the equation of the curve. Now, for the equation of the tangent, we need the gradient of the curve at that point. It can be found as,
f(x) = x3 – x + 3
f′(x) = 3x2 – 1
Then, f′(x = 1) = 3.(1)2 – 1 = 2 = m
The tangent would be the straight line passing through (1,3) with slope = 2.
(y – 3) = 2(x – 1)
y = 2x + 1
2x – y + 1 = 0
b) The normal would pass through the point (1,3) and its slope n would be given by,
n = -(1/m) = -(1/2) = -0.5
Equation of the normal:
(y – 3) = -0.5(x – 1)
2y = -x + 7
x – 2y – 7 = 0
Question 2: Explain the difference between a tangent and a normal?
Answer: A tangent refers to a straight line whose extension takes place from a point on a curve, with a gradient equal to the curve’s gradient existing at that particular point. A normal, in contrast, refers to a straight line whose extension takes place from a curve’s point such that it is perpendicular to the point’s tangent.
Question 3: Explain how can one find the tangent?
Answer: One can find the tangent by the following steps:
- Sketch the tangent line and the function.
- Take the first derivative in order to find the equation for the tangent line’s slope.
- Enter the x value belonging to the point under investigation.
- In point-slope form, write the equation of the tangent line.
- Finally, do confirmation of the equation on the graph.
Question 4: Can we say that the gradient is the same as a slope?
Answer: Gradient refers to the degree of a graph’s steepness at any point. Slope refers to the graph’s gradient at any point. So, one can say that both are the same.
Question 5: Name the four kinds of slopes?
Answer: The four kinds of slopes are zero, undefined, positive, and negative.
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