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5.4 Tangents and normalsIB Maths: Applications and Interpretation SL: Revision notes

Section 1

Tangents, normals and gradients

A tangent to a curve at a point is the straight line that touches the curve there and has the same gradient as the curve. Its gradient is the value of the derivative at that point: mtangent=f′(a)m_{\text{tangent}}=f'(a) at x=ax=a. A normal is the straight line through the same point that is perpendicular to the tangent. Perpendicular gradients multiply to −1-1, so mnormal=−1f′(a).m_{\text{normal}}=-\frac{1}{f'(a)}. Take the reciprocal and change the sign. For example, a tangent gradient of 88 gives a normal gradient of −18-\frac18; a tangent gradient of −23-\frac23 gives a normal gradient of 32\frac32. If the tangent is horizontal (f′(a)=0f'(a)=0), the normal is the vertical line x=ax=a.

Key termstangentnormalperpendicular
Common mistake

Forgetting either the reciprocal or the negative sign when finding the normal gradient.

Section 2

Equation of a tangent

Steps:

  1. Find yy at the given xx to get the point (x1,y1)(x_1,y_1) (if not given).
  2. Differentiate and substitute x1x_1 to get the gradient mm.
  3. Use y−y1=m(x−x1)y-y_1=m(x-x_1) and rearrange as the question asks (for example y=mx+cy=mx+c or ax+by+d=0ax+by+d=0). Example: y=x3−4x+1y=x^3-4x+1 at P(2,1)P(2,1). dydx=3x2−4=8\frac{dy}{dx}=3x^2-4=8 at x=2x=2. Tangent: y−1=8(x−2)y-1=8(x-2), so y=8x−15y=8x-15.
Key termspoint-gradient form
Common mistake

Using the yy-value as the gradient, or substituting into yy instead of dydx\frac{dy}{dx} to get the gradient.

Exam tip

Check your tangent: substituting the point into your equation must give a true statement.

Section 3

Equation of a normal

Use the same point, but the perpendicular gradient. For the curve above, the normal at P(2,1)P(2,1) has gradient −18-\frac18: y−1=−18(x−2)y-1=-\frac18(x-2), so y=−18x+54y=-\frac18x+\frac54, or x+8y−10=0x+8y-10=0. To find where a tangent or normal meets an axis, set y=0y=0 for the xx-axis or x=0x=0 for the yy-axis. The length of a normal between two points comes from Pythagoras: (Δx)2+(Δy)2\sqrt{(\Delta x)^2+(\Delta y)^2}.

Exam tip

Write the tangent gradient first, then the normal gradient, so you do not mix them up when you substitute.

Section 4

Points with a given gradient

To find where the tangent has gradient kk, solve f′(x)=kf'(x)=k, then find yy for each solution. A tangent is parallel to a line with gradient kk when f′(x)=kf'(x)=k. Example: for y=2x2−5x+3y=2x^2-5x+3 the tangent is parallel to y=3x+1y=3x+1 when 4x−5=34x-5=3, so x=2x=2 and the point is (2,1)(2,1). If f′(x)=kf'(x)=k is a quadratic equation, there may be two solutions, so give both points. Where f′(x)=0f'(x)=0 the tangent is horizontal.

Key termsparallel
Common mistake

Finding the xx-value where the gradient is kk but stopping before finding the yy-coordinate that the question asks for.

Section 5

Using technology

A GDC can graph y=f(x)y=f(x) and draw the tangent at a chosen point, then display the tangent's equation. It can also evaluate the gradient at a point numerically, solve f′(x)=kf'(x)=k, and find the intersection of a tangent or normal with the curve or an axis. Use technology to check an analytic answer or when the numbers are awkward, and quote results to three significant figures unless exact values are asked for. For 'show that' or 'find the exact' questions, show the algebra. In context questions, state what the tangent or normal represents, for example the slope of a hillside or the direction of a tunnel, and include units for lengths.

Exam tip

Use the GDC tangent tool to check that your gradient and equation are correct before moving on.

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Exam questions on 5.4 Tangents and normals

  1. The curve CC has equation y=x3−4x+1y=x^3-4x+1 and passes through the point P(2,1)P(2,1).
    Find the equation of the tangent to CC at PP, giving your answer in the form y=mx+cy=mx+c.2 marks
  2. A curve has equation y=2x2−5x+3y=2x^2-5x+3.
    Find the equation of the tangent to the curve at the point where x=3x=3. Give your answer in the form ax+by+d=0ax+by+d=0, where aa, bb and dd are integers.2 marks
  3. A curve has equation y=x3−3x2+2x+4y=x^3-3x^2+2x+4. The point AA on the curve has xx-coordinate 33.
    Find the equation of the tangent to the curve at AA.3 marks
See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).