Tangents and Normals to a Curve
Find the equation of the tangent at a point on a curve from the derivative there, and the normal from the perpendicular gradient.
What a learner can do afterwards
- Find the tangent to y = x² - 3x at the point where x = 2
- Find the normal at the same point and check the gradients multiply to -1
- Find the point where a given tangent meets the x-axis
1 · Read
Last stop you read dy/dx as a rate with units. At a single point it is still the gradient of the tangent, and now that number goes back into a line. Now build the line itself. On the Ridgeway ramp a board leaves the surface at the lip and carries straight on along the tangent there, so the crew needs that line's equation to work out where the board comes down. Two things fix any line, and you can produce both: a point it passes through, and its gradient. The curve gives you the first and the derivative gives you the second.
So the recipe has three steps and no new machinery. First the point: put the x value into the curve's own rule to get y. Second the gradient: put the same x into dy/dx. Third the line: feed both into y - y₁ = m(x - x₁), the point-gradient form from the coordinate work. The normal is the line through the same point at right angles to the tangent, so its gradient is the negative reciprocal, -1/m. Everything else about it is identical. Some questions hand you the gradient instead, and then you put dy/dx equal to that number and solve for x.
Find the tangent to y = x² - 3x at x = 2. The point comes from the curve: 2² - 3 times 2 is 4 - 6, which is -2, so the point is (2, -2). The gradient comes from the derivative: dy/dx = 2x - 3, and at x = 2 that is 4 - 3 = 1. Now the point-gradient form: y - (-2) = 1(x - 2), which is y + 2 = x - 2, so y = x - 4. Check the point on the answer: at x = 2 it gives -2, which matches.
Now the normal at that same point. The tangent gradient was 1, so the normal gradient is -1/1, which is -1. The point has not changed, so y + 2 = -1(x - 2), which gives y = -x. Check it: (2, -2) satisfies y = -x, and 1 times -1 = -1, the right-angle test from the coordinate work. When the flip gives a fraction, clear it: multiplying y - 8 = -(1/7)(x - 2) by 7 gives 7y - 56 = -x + 2, so x + 7y - 58 = 0. A gradient of 0 has no flip, so the normal there is vertical.
Once you have it, a tangent is just a line, so every coordinate move applies. The tangent above was y = x - 4. For the x-axis crossing set y = 0: 0 = x - 4, so x = 4 and the crossing is (4, 0), which on the ramp is where the board comes back down to the ground. For the y-axis crossing set x = 0, giving (0, -4). The one thing to keep straight all the way through is which rule feeds which slot. The curve gives heights, the derivative gives gradients, and no answer ever needs them swapped.
To build a tangent at a given x, take the point from the curve's own rule and the gradient from dy/dx, then put both into y - y₁ = m(x - x₁). The normal runs through the same point with gradient -1/m, so the two gradients multiply to -1. Once either line is written down it behaves like any other line: set y = 0 for its x-axis crossing and x = 0 for its y-axis crossing.
2 · Watch
Take it off screen
Where it sits
24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.