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Differentiation I

Gradients, tangents and normals

Gradients, tangents and normals

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Summary

Gradients, tangents and normals

In a nutshell

The gradient of a curve at a particular point on it can be found using the derivative of the equation of the curve. Moreover, equations for tangents to curves and also normals to tangents can be found using differentiation. In particular, the gradient of a tangent to a curve is the gradient of the curve at the point the tangent meets it. Normals are perpendicular to tangents, so their gradients are given by the negative reciprocal of the gradient of the tangent.



Gradients of curves

Curves do not have constant gradients. You use the derivative of the equation of the curve and substitute the xxx-coordinate of some point to find the gradient of the curve at that particular point.


Example 1

Calculate the gradient of the curve y=4x2+3x−5y=4x^2+3x-5y=4x2+3x−5 at the point (1,2)(1,2)(1,2).


First differentiate the equation of the curve with respect to xxx.

dydx=8x+3\dfrac{\text{d}y}{\text{d}x}=8x+3dxdy​=8x+3​​


Now substitute the xxx-coordinate of the given point into this equation. When x=1x=1x=1:

dydx=8\dfrac{\text{d}y}{\text{d}x}=8dxdy​=8​​


Therefore, at the point (1,2)(1,2)(1,2) of the curve given by the equation y=4x2+3x−5y=4x^2+3x-5y=4x2+3x−5 the gradient is 8‾\underline{8}8​.



Tangents

A tangent is a straight line that touches a curve at a point. It may cross the curve at some other location, but it only touches the point of focus. Since a tangent is a straight line, it has the equation y=mx+cy=mx+cy=mx+c. In particular, the gradient of the tangent, mmm, is equal to the gradient of the curve at the point where the tangent touches it. It's important to realise that the point where the tangent and curve meet is a common point between them. Hence, it can be used to find the equation of the tangent.


PROCEDURE

1.1.1.​

Differentiate the equation of the curve.

2.2.2.​

Substitute the xxx-coordinate of the given point into the equation for the derivative found in the previous step. This gives the gradient of the tangent, mmm, at the given point.

3.3.3.​

Use mmm in the equation of the straight line: y−y1=m(x−x1)y-y_1=m(x-x_1)y−y1​=m(x−x1​). 

4.4.4.​

Insert the coordinates of the given point into the equation for the tangent from the previous step and express in the form y=mx+cy=mx+cy=mx+c. Now you have the equation of the tangent.


Example 2

Find the equation of the tangent to the curve y=2x3−4xy=2x^3-4xy=2x3−4x at the point (−1,2)(-1,2)(−1,2).


The gradient of the tangent is the same as the gradient of the curve at (−1,2)(-1,2)(−1,2), so start by differentiating the equation of the curve.

dydx=6x2−4\dfrac{\text{d}y}{\text{d}x}=6x^2-4dxdy​=6x2−4​​


Insert the xxx-coordinate from the given point into this equation. Hence, when x=−1x=-1x=−1, the gradient of the curve is:

dydx=6(−1)2−4=2\dfrac{\text{d}y}{\text{d}x}=6(-1)^2-4=2dxdy​=6(−1)2−4=2​​


Thus, you know that the tangent has a gradient of 222. So far, it's equation can be given by y−y1=2(x−x1)y-y_1=2(x-x_1)y−y1​=2(x−x1​). Since the point (−1,2)(-1,2)(−1,2) is by definition on this line, insert its coordinates.

y−2=2(x+1) y−2=2x+2 y=2x+4y-2=2(x+1)\\\space\\y-2=2x+2\\\space\\y=2x+4y−2=2(x+1) y−2=2x+2 y=2x+4​​


You can conclude that the equation of the tangent is given by:

 y=2x+4‾\underline{y=2x+4}y=2x+4​.



Normals

Normals are lines that are perpendicular (at right angles) to some other line. In this context, normals are lines that are perpendicular to tangents to curves. In other words, normals are perpendicular to curves at a given point. To find the gradient of a perpendicular line, take the negative reciprocal of the gradient of the original line. This means to do negative one over the original gradient.


Example 3

What is the negative reciprocal of 222?


You do negative one over two. So the negative reciprocal of 222 is: 

−12‾\underline{-\dfrac12}−21​​​​


Example 4

What is the negative reciprocal of mmm?


You do negative one over mmm​. So the negative reciprocal of mmm is: 

−1m‾\underline{-\dfrac1m}−m1​​​​


Example 5

Find the equation for the normal to the curve y=x2y=x^2y=x2 at the point (3,9)(3,9)(3,9).


First find the gradient of the curve at (3,9)(3,9)(3,9). Do this by first differentiating the equation of the curve.

dydx=2x\dfrac{\text{d}y}{\text{d}x}=2xdxdy​=2x​​


Thus, the gradient of the curve at (3,9)(3,9)(3,9) is 2(3)=62(3)=62(3)=6.


The gradient of the normal is the negative reciprocal of this value: −16-\dfrac16−61​. So the normal has equation y−y1=−16(x−x1)y-y_1=-\dfrac16(x-x_1)y−y1​=−61​(x−x1​). Use the given point to find the equation of the normal.

y−9=−16(x−3) y−9=−x6+12 y=−x6+192y-9=-\dfrac16(x-3)\\\space\\y-9=-\dfrac{x}6+\dfrac12\\\space\\y=-\dfrac{x}6+\dfrac{19}2y−9=−61​(x−3) y−9=−6x​+21​ y=−6x​+219​​​


Therefore, the gradient of the normal to the curve at the given point is:

 y=−16x+192‾\underline{y=-\frac16x+\frac{19}2}y=−61​x+219​​



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FAQs - Frequently Asked Questions

What are normals to curves?

Normals are lines that are perpendicular (at right angles) to some other line. In this context, normals are lines that are perpendicular to tangents to curves.

What is a tangent to a curve?

A tangent is a straight line that touches a curve at a point. It may cross the curve at some other location, but at the point of focus, it only touches.

How do you find the gradient of a curve at a point?

The gradient of a curve at a particular point on it can be found using the derivative of the equation of the curve and substituting the x coordinate into the derivative equation.

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