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 a point and tangent gradient on a curve, then construct the tangent equation
- Construct a normal; check the gradient product when both gradients are finite, and handle horizontal tangents with vertical normals
- Find axis crossings of tangent and normal lines from their equations
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A derivative describes a rate of change, with units of output per unit of input. For y against x, a finite dy/dx at a point is the tangent's gradient. A point and a gradient fix a nonvertical line: the curve supplies the point and its derivative supplies the gradient. In the Ridgeway ramp's straight-line model, a board continues along the tangent from the lip. The crew uses its equation to find where the line reaches the ground.
Put x into the curve's rule to get y. Differentiate and put the same x into dy/dx to get finite gradient m. Use y-y₁=m(x-x₁). If a tangent gradient is supplied, use it directly. If asked where a curve has a chosen gradient, set dy/dx equal to that value and solve for x. A normal passes through the same point at right angles to the tangent. For finite m≠0, its gradient is -1/m. For m=0, its equation is x=x₁ and its gradient is undefined.
For y=x²-3x at x=2, the curve gives y=4-6=-2, so the point is (2,-2). Differentiate: dy/dx=2x-3, giving m=1 there. Thus y+2=1(x-2), or y=x-4. Check both the point and gradient: x=2 gives y=-2, and the line's gradient is 1. For reciprocal or root curves, first rewrite the function as powers, then differentiate and use the same point-gradient method.
Point from the curve, gradient from the derivative, then one substitution.
A normal meets the tangent at right angles through the same point. With finite tangent gradient m≠0, its gradient is -1/m and the product is -1. At (2,-2), m=1 gives y+2=-(x-2), so y=-x. For a fractional example, y-8=-(1/7)(x-2) becomes x+7y-58=0 after multiplying by 7. Opposite signs alone do not prove perpendicularity: 7 times -7 is -49, not -1. If m=0, the normal is vertical, x=x₁, with no gradient.
For a finite nonzero tangent gradient, take the negative reciprocal. A horizontal tangent has a vertical normal.
For an axis crossing, first build the line. Get (x₁,y₁) from the curve and finite tangent gradient m from dy/dx. Use y-y₁=m(x-x₁) for the tangent, or y-y₁=-(1/m)(x-x₁) for the normal when m≠0. Then set y=0 for an x-axis crossing or x=0 for a y-axis crossing and solve. The tangent y=x-4 meets the axes at (4,0) and (0,-4). These are coordinate crossings, not a claim about a board moving downwards.
Build the tangent or normal first. Set y=0 for the x-axis crossing and x=0 for the y-axis crossing.
Point from curve; m from dy/dx. Use y-y₁=m(x-x₁). For finite m≠0, normal gradient -1/m; product -1. For m=0, normal x=x₁. Axis crossings: y=0 for x-axis, x=0 for y-axis.
The three readings and six practice questions were written for this dataset. The linked videos are Khan Academy's "Connecting the derivative and tangent line equations" and "Finding equations of tangents to polynomials".
More to read
Building a tangent from its point and gradient
At a chosen point, a tangent follows the curve's direction. Its gradient equals the derivative evaluated there.
First substitute the given x into the original function to find the point's y-coordinate. Differentiate the function, then substitute that x into the derivative for gradient m. Write the tangent using y - y₁ = m(x - x₁).
For y = x^2 - 3x, choosing x = 2 gives y = -2. The derivative is 2x - 3, so m = 1 there. Substitution gives y + 2 = x - 2, or y = x - 4.
You can also locate the intercepts. Using y = 0 finds the x-axis crossing; using x = 0 finds the y-axis crossing.
Estimating a derivative with a drawn tangent
A graph can help you estimate a derivative. Choose a location on the curve and use a ruler to sketch a line matching its direction there. This is an approximate tangent.
Take two well-separated points on your drawn line. Divide the change in their y-coordinates by the change in their x-coordinates to estimate its gradient. That number estimates the derivative at the chosen location.
A tangent with greater steepness indicates a larger magnitude of change. A horizontal tangent gives a derivative of zero. Comparing this graphical estimate with an algebraic answer helps you check whether the answer makes sense.
Choosing a tangent or a normal
Start by identifying which line the question asks for. A tangent has the curve's derivative as its gradient at the contact point. A normal passes through that point at a right angle to the tangent.
When the tangent gradient m is finite and nonzero, the normal gradient is -1/m. Their product is -1. If the tangent is horizontal, m=0 and the normal is vertical. Write that normal as x equal to the contact point's x-coordinate; it has no defined gradient.
For a nonvertical line, substitute its point and gradient into y-y₁=m(x-x₁). Check the point by substitution, then check the gradient separately. Passing through the contact point does not by itself prove that a line is the tangent or normal.
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Where it sits
30 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.