How Do You Find the Slope of a Line Normal to a Curve

The slope of a function can be calculated at any point along its length. The function will have the same slope at all points if it is a straight line. However, the slope of any function that isn’t a straight line will change as the function’s value changes.

We can use the derivative of a function and then evaluate it at the location we’re interested in to get the slope of a function at a specific point. This not only provides us with the slope of the function at that location but also the slope of the tangent line to that function.

At the point where the tangent line intersects the function, the normal line is perpendicular to the tangent line. This indicates that if the tangent line’s slope is mm, the normal line’s slope is the negative reciprocal of mm, or -1/m1/m.

To obtain the answer to the question on how to find the slope of a line normal to a curve, simply follow the instructions below.

1. Take the original function’s derivative and evaluate it at the specified position. This is the tangent line’s slope, which we’ll call mm.
2. Find the negative reciprocal of mm, or -1/m1/m in other terms. This is the normal line’s slope, which we’ll name nn. So n=-1/mn=−1/m.
3. For the equation of the line, plug nn and the provided point into the point-slope formula, (y-y1)=n(x-x1)(y−y​1​​)=n(x−x​1​​).
4. Solve for yy to simplify the equation.

Tangent and normal lines

A tangent line is a line that intersects a curve at only one place. It is important to grasp the basic differentiation rules of differential calculus to discover the derivative function f'(x) of the original function f to determine its slope (x). The slope of the tangent line at a given location is equal to the value of f'(x). Once you have the slope, you can use the point-slope formula to obtain the tangent line’s equation: (y – y1) = (m(x – x1)).

The derivative of a function can be used to solve a variety of calculus problems. Curve drawing, addressing maximum and minimum problems, distance, velocity, and acceleration difficulties, solving associated rate problems, and approximating function values are all possible applications.

The slope of the tangent line at a location is the derivative of a function at that point. At the point of tangency, the normal line is defined as the line that is perpendicular to the tangent line. Because the slopes of perpendicular lines (none of which are vertical) are negative reciprocals of each other, the normal line to the graph of f(x) is −1/ f'(x).

How do you find the slope of a line normal to a curve?

The equation of a line in analytic geometry is y = mx + b, where m is the slope and b is the y-intercept. The negative reciprocal, or -1/m, of any line perpendicular to a line with slope m is the slope of any line perpendicular to a line with slope m. Because we don’t always use the y-intercept, a different type of line equation is usually used to find the slope of a line normal to a curve.

Solved examples

Example 1: Find the slope of the curve y = (1 + x) sin x at x = π/4.

Solution: 

Given equation is y = (1 + x) sin x.

Apply the derivative on the above equation.

dy/dx = (1 + x) cos x + sin x

[Using product rule]

Now, slope at x = π/4

dy/dx = (1 + π/4) cos (π/4) + sin(π/4)

= 1/√2 . (2 + π/4) or (2 + π/4)/√2

Which is the slope of the given curve.

Problem: Find the slope of the normal to the curve x = a cos3 θ, y = a sin3 θ at θ = π/4.

Solution:

It is given that x = a cos3 θ, y = a sin3 θ

dx/dθ = 3a cos2 θ * (-sin θ) = -3a cos2 θ * sin θ

dy/dθ = 3a sin2 θ * cos θ

Now, dy/dx = (dy/dθ)/( dy/dθ)

= (3a sin2 θ * cos θ)/( -3a cos2 θ * sin θ)

= – sin θ /cos θ

= – tan θ

Therefore, the slope of the tangent at θ = π/4 is given by,

dy/dx]θ = π/4 = -tan π/4 = -1

Now, the slope of the normal at θ = π/4 is given as

= -1/[dy/dx]θ = π/4

= -1/(-1)

= 1

Conclusion

The tangent, in general, describes the instantaneous rate of change of a function at a given position. As a result, the slope of the tangent at a given position is equal to the function’s derivative at that same location.

As a result, as x approaches 0, the slope of the tangent is defined as the limit of Δy/Δx, as Δx. The slope of the tangent at a given position is determined by the value at that location on the provided function.