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AI SL · 12.9

Differentiation

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12.9.8

Equation of a tangent and of a normal

12.9.8

Equation of a tangent and of a normal

Target: I can find the gradient and the y-coordinate of a curve at a given x, and use \(y-y_1=m(x-x_1)\) to write the equation of the tangent. (+3 more)

By the end of this section

Key points

The tangent

  • The tangent to \(y=f(x)\) at \(x=a\) is the straight line through the point \((a, f(a))\) with gradient \(f'(a)\).
  • Equation of the tangent: \(y-y_1=m(x-x_1)\), where \(m=f'(a)\), \(x_1=a\) and \(y_1=f(a)\).
  • Two numbers are needed: the gradient from \(\frac{dy}{dx}\), and the y-coordinate from the curve itself. Do not mix them up.

The normal

  • The normal at a point is the line through that point perpendicular to the tangent.
  • Perpendicular gradients multiply to \(-1\), so \(m_{normal}=-\frac{1}{f'(a)}\): turn the gradient upside down and change its sign.
  • The normal passes through the same point \((a, f(a))\) as the tangent. Only the gradient changes.
  • If \(f'(a)=0\) the tangent is horizontal and the normal is the vertical line \(x=a\).

Giving the answer in the right form

  • Read the form in the question. \(y=mx+c\), \(ax+by=c\) and \(ax+by+c=0\) all describe the same line.
  • With integers a, b, c, multiply through to clear every fraction. Write the line so that a is positive.
  • Check by putting the point back into your final equation.

With the calculator

  • When a curve has \(\sqrt{x}\), \(\ln x\), \(e^x\) or a product such as \(x\ln x\), the calculator gives the gradient at the point.
  • Keep the full values from the calculator until the last line, then give m and c to 3 significant figures.

On the calculator

Gradient of a curve at a point

Menu > Calc > Derivative at a point

  • Type the function and the value of x.
  • The number shown is the gradient of the tangent at that x.
  • Find the y-coordinate by putting the same x into the function.

Screenshot and video to come

Building the line from the readings

  • Write \(y-y_1=m(x-x_1)\) with the full calculator values for m and \(y_1\).
  • Make y the subject, then round m and c to 3 significant figures.

Screenshot and video to come

Explore: the tangent at a point

Drag the orange point P along the curve. The orange line is the tangent at P. Its gradient is f′(x) at that point.


Curve
Point
Gradient at P
Tangent (orange)

The two axes have different scales, so the normal does not always look perpendicular on the screen.

Drag P along the curve.

  1. On the quadratic, move P to the lowest point. What is the gradient there? What does the tangent look like?
  2. On the cubic, where is the gradient negative? What is the curve doing there? Why?
  3. Show the normal. Multiply the gradient of the tangent by the gradient of the normal. What do you get? Why does a horizontal tangent break the rule?
  4. On the curve y = 4x + 25/x, what happens to the gradient as x gets larger? Why?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Tangent to a curve with a reciprocal term

Find the equation of the tangent to the curve \(y=x^3-\frac{3}{x}+1\) at the point where \(x=1\). Give your answer in the form \(ax+by=c\).

Working
  1. Differentiate

    \(\frac{dy}{dx}=3x^{2}+3x^{-2}\)

    Write \(-\frac{3}{x}\) as \(-3x^{-1}\), then lower each power by 1.

  2. Gradient and point

    When \(x=1\): \(m=6\) and \(y=-1\)

    The gradient m is the value of the derivative at \(x=1\). Put \(x=1\) into the curve for the y-coordinate.

  3. Line through the point

    \(y+1=6(x-1)\)

    Use \(y-y_1=m(x-x_1)\) with the point \((1, -1)\).

  4. Required form

    \(6x-y=7\)

    Collect the x and y terms on the left and the number on the right.

Video walkthrough coming
Normal in general form

Find the equation of the normal to the curve \(y=2x^2+\frac{8}{x}-1\) at the point where \(x=2\). Give your answer in the form \(ax+by+c=0\), where a, b and c are integers.

Working
  1. Tangent gradient and point

    \(\frac{dy}{dx}=4x-8x^{-2}\); When \(x=2\): \(m=6\) and \(y=11\)

    Differentiate, then substitute \(x=2\).

  2. Normal gradient

    Normal gradient: \(m=-\frac{1}{6}\)

    The normal is perpendicular to the tangent: turn the gradient upside down and change the sign.

  3. Line through the point

    \(y-11=-\frac{1}{6}(x-2)\)

    Same point \((2, 11)\), new gradient.

  4. Clear the fraction

    \(x+6y-68=0\)

    Multiply by 6 to remove the fraction, then collect everything on the left. The integers are 1, 6 and the constant.

Video walkthrough coming
Normal to a quartic

Find the equation of the normal to the curve \(y=x^4-3x^2+2\) at the point where \(x=-1\). Give your answer in the form \(y=mx+c\).

Working
  1. Differentiate

    \(\frac{dy}{dx}=4x^{3}-6x\)

    Lower each power by 1.

  2. Tangent gradient and point

    When \(x=-1\): \(m=2\) and \(y=0\)

    Watch the signs with \(x=-1\): odd powers stay negative.

  3. Normal gradient

    Normal gradient: \(m=-\frac{1}{2}\)

    Perpendicular gradients multiply to \(-1\).

  4. Equation

    \(y=-\frac{1}{2}(x+1)\); \(y=-\frac{1}{2}x-\frac{1}{2}\)

    Expand and make y the subject.

Video walkthrough coming
Tangent with the calculator

Use technology to find the equation of the tangent to the curve \(y=x\ln x\) at the point where \(x=2\). Give your answer in the form \(y=mx+c\), with m and c to 3 significant figures.

Working
  1. Gradient

    From the calculator: \(\frac{d}{dx}(x\ln x)\) at \(x=2\) is \(1.693\)

    You do not need to differentiate \(x\ln x\) by hand. Use the derivative at a point.

  2. The point

    \(y=2\ln 2=1.386\)

    Find y from the curve, with the calculator.

  3. Line

    \(y-1.386=1.693(x-2)\)

    Keep the full calculator values until the last line.

  4. Answer

    \(y=1.69x-2.00\)

    Give m and c to 3 significant figures.

Video walkthrough coming