IB Maths AA 5.2 Notes

Tangents & normals

Derivatives are not only used to find gradients. They are also used to find equations of tangents and normals, to describe the shape of a graph, and to locate maximum points, minimum points and points of inflexion.

A tangent is a straight line that touches a curve at a given point and has the same gradient as the curve at that point. If the curve is $y=f(x)$, then the gradient of the tangent at $x=a$ is:

$m = f'(a)$

To find the equation of a tangent:

1. Find the derivative $f'(x)$.
2. Substitute the given $x$-value to find the gradient.
3. Find the corresponding point on the curve.
4. Use $y-y_1=m(x-x_1)$ to find the equation.

Find the equation of the tangent to $y=x^2+3x-1$ at $x=2$.

Differentiate: $f'(x)=2x+3$.

At $x=2$, $f'(2)=2(2)+3=7$, so the gradient is $7$.

Find the point on the curve: $y=2^2+3(2)-1=9$, so the point is $(2,9)$.

Use point-gradient form: $y-9=7(x-2)$.

Simplify: $y-9=7x-14$, so $y=7x-5$.

A normal is a straight line perpendicular to the tangent at the point of contact. If the gradient of the tangent is $m$, then the gradient of the normal is

$-\frac{1}{m}$

To find the equation of a normal:

1. Find the gradient of the tangent $f'(x)$.
2. Take the negative reciprocal $-\frac{1}{m}$.
3. Use the same point on the curve.
4. Use $y-y_1=m(x-x_1)$.

Find the equation of the normal to $y=x^2+3x-1$ at $x=2$.

From the previous example, the tangent gradient is $7$, so the normal gradient is $-\frac{1}{7}$.
The point is still $(2,9)$.

So the normal is $y-9=-\frac{1}{7}(x-2)$

Find the tangent and normal to $y=x^3-2x$ at $x=1$.

Differentiate: $f'(x)=3x^2-2$.

At $x=1$, $f'(1)=3(1)^2-2=1$, so the tangent gradient is $1$.

The point on the curve is $(1,-1)$ because $1^3-2(1)=-1$.

Tangent: $y+1=1(x-1)$, so $y=x-2$.

Normal gradient is $-1$.

Normal: $y+1=-1(x-1)$, so $y=-x$.