1. Tangents and Normals
by M. Bourne
We often need to find tangents and normals to curves when we are analysing forces acting on a moving body.
A tangent to a curve is a line that touches the curve at one point and has the same slope as the curve at that point.
A normal to a curve is a line perpendicular to a tangent to the curve.
the curve
the curve
Graph showing the tangent and the normal to a curve at a point.
Note 1: As we discussed before (in Slope of a Tangent to a Curve), we can find the slope of a tangent at any point (x, y) using `dy/dx`.
Note 2: To find the equation of a normal, recall the condition for two lines with slopes m1and m2 to be perpendicular (see Perpendicular Lines):
m1 × m2 = −1
Applications
A car has skidded while rounding a corner, tangent to the double yellow lines curve.
Tangent:
- If we are traveling in a car around a corner and we drive over something slippery on the road (like oil, ice, water or loose gravel) and our car starts to skid, it will continue in a direction tangent to the curve.
- Likewise, if we hold a ball and swing it around in a circular motion then let go, it will fly off in a tangent to the circle of motion.
The spokes of a bicycle wheel are normal to the rim.
Normal:
- When you are going fast around a circular track in a car, the force that you feel pushing you outwards is normal to the curve of the road. Interestingly, the force that is making you go around that corner is actually directed towards the center of the circle, normal to the circle.
- The spokes of a wheel are placed normal to the circular shape of the wheel at each point where the spoke connects with the center.
Examples
Need Graph Paper?
1. Find the gradient of
(i) the tangent (ii) the normal
to the curve y = x3 − 2x2 + 5 at the point `(2,5)`.
Answer
`dy/dx=3x^2-4x`
The slope of the tangent is
`m_1=[dy/dx] _(x=2)`
`=3(2)^2-4(2)`
`=12-8`
`=4`
The slope of the normal is found using m1 × m2 = −1
`m_2=-1/4`
2. Find the equation of (i) the tangent and (ii) the normal in the above example.
Answer
We use y − y1 = m(x − x1), with x1 = 2, y1 = 5
(i) The tangent has slope `4`, so we have:
`y-5=4(x-2)`
gives
`y=4x-3`
or
4x − y − 3 = 0
(ii) Now for the normal to the curve. Since the tangent has slope `4`, we have the slope of the normal `m=-1/4`
So we substitute as follows:
`y-5=-1/4(x-2)`
gives
`y=-1/4x+5 1/2`
or
x + 4y − 22 = 0
3. Sketch the curve and the normal in the above example.
Answer
Here's the graph of the tangent and normal to the curve at `x=2`.
