Tangents and normals Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Tangents and normals questions. See exactly how to solve problems on equation of a tangent, differentiation, gradient at a point, cubic curves.

equation of a tangentdifferentiationgradient at a pointcubic curvesquadratic curvesequation of a normal
A-Level70 questionsStep-by-step solutions
Question 1
3 markseasy
Find the equation of the tangent to the curve y=x2y = x^2 at the point where x=3x = 3.

Worked solution

  1. Differentiate the curve to get the gradient function

    dydx=2x\frac{dy}{dx} = 2 x

    Differentiating tells us the gradient at any point on the curve. Remember the power rule from the differentiation topic: multiply by the power, then subtract one from the power.

  2. Find the yy-coordinate of the point of contact

    x=3  ⇒  y=9x = 3 \;\Rightarrow\; y = 9

    Put the xx-value into the original curve to find where the tangent touches. This gives the exact point the line passes through.

  3. Work out the gradient at that point

    dydx∣x=3=6\left.\frac{dy}{dx}\right|_{x=3} = 6

    Substitute the xx-value into the gradient function. This number is the gradient of the tangent line.

  4. Substitute the point and gradient

    y−9=6(x−3)y - 9 = 6(x - 3)

    Put the gradient and the point into the straight-line formula. We are almost there — just tidy it up.

  5. Simplify to the form y=mx+cy = mx + c

    y=6x−9y = 6 x - 9

    Expand the bracket and collect terms to write the tangent neatly. This is the equation of the tangent.

Answer
y=6x−9y = 6 x - 9
Question 2
2 markseasy
Find the gradient of the curve y=x2y = x^2 at the point where x=−2x = -2.

Worked solution

  1. Differentiate the curve

    dydx=2x\frac{dy}{dx} = 2 x

    The derivative is the gradient function. It tells us the steepness of the curve at any xx.

  2. Substitute the given xx-value

    dydx∣x=−2=−4\left.\frac{dy}{dx}\right|_{x=-2} = -4

    Put the xx-value into the gradient function to get the gradient at that exact point.

  3. State the gradient

    gradient=−4\text{gradient} = -4

    This single number is the gradient of the curve (and of the tangent) at the given point.

Answer
−4-4
Question 3
3 markseasy
Find the equation of the tangent to the curve y=x3y = x^3 at the point where x=1x = 1.

Worked solution

  1. Differentiate the curve to get the gradient function

    dydx=3x2\frac{dy}{dx} = 3 x^{2}

    Differentiating tells us the gradient at any point on the curve. Remember the power rule from the differentiation topic: multiply by the power, then subtract one from the power.

  2. Find the yy-coordinate of the point of contact

    x=1  ⇒  y=1x = 1 \;\Rightarrow\; y = 1

    Put the xx-value into the original curve to find where the tangent touches. This gives the exact point the line passes through.

  3. Work out the gradient at that point

    dydx∣x=1=3\left.\frac{dy}{dx}\right|_{x=1} = 3

    Substitute the xx-value into the gradient function. This number is the gradient of the tangent line.

  4. Substitute the point and gradient

    y−1=3(x−1)y - 1 = 3(x - 1)

    Put the gradient and the point into the straight-line formula. We are almost there — just tidy it up.

  5. Simplify to the form y=mx+cy = mx + c

    y=3x−2y = 3 x - 2

    Expand the bracket and collect terms to write the tangent neatly. This is the equation of the tangent.

Answer
y=3x−2y = 3 x - 2
Question 4
3 markseasy
Find the equation of the tangent to the curve y=x2−4x+1y = x^2 - 4x + 1 at the point where x=1x = 1.

Worked solution

  1. Differentiate the curve to get the gradient function

    dydx=2x−4\frac{dy}{dx} = 2 x - 4

    Differentiating tells us the gradient at any point on the curve. Remember the power rule from the differentiation topic: multiply by the power, then subtract one from the power.

  2. Find the yy-coordinate of the point of contact

    x=1  ⇒  y=−2x = 1 \;\Rightarrow\; y = -2

    Put the xx-value into the original curve to find where the tangent touches. This gives the exact point the line passes through.

  3. Work out the gradient at that point

    dydx∣x=1=−2\left.\frac{dy}{dx}\right|_{x=1} = -2

    Substitute the xx-value into the gradient function. This number is the gradient of the tangent line.

  4. Substitute the point and gradient

    y+2=−2(x−1)y + 2 = -2(x - 1)

    Put the gradient and the point into the straight-line formula. We are almost there — just tidy it up.

  5. Simplify to the form y=mx+cy = mx + c

    y=−2xy = - 2 x

    Expand the bracket and collect terms to write the tangent neatly. This is the equation of the tangent.

Answer
y=−2xy = - 2 x
Question 5
3 markseasy
Find the equation of the normal to the curve y=x2y = x^2 at the point where x=2x = 2.

Worked solution

  1. Differentiate the curve to get the gradient function

    dydx=2x\frac{dy}{dx} = 2 x

    Differentiating tells us the gradient at any point on the curve. Remember the power rule from the differentiation topic: multiply by the power, then subtract one from the power.

  2. Find the yy-coordinate of the point of contact

    x=2  ⇒  y=4x = 2 \;\Rightarrow\; y = 4

    Put the xx-value into the original curve to find where the tangent touches. This gives the exact point the line passes through.

  3. Work out the gradient at that point

    dydx∣x=2=4\left.\frac{dy}{dx}\right|_{x=2} = 4

    Substitute the xx-value into the gradient function. This number is the gradient of the tangent line.

  4. Find the gradient of the normal

    mn=−14=−14m_n = -\dfrac{1}{4} = - \frac{1}{4}

    The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient. Flip the fraction and change the sign.

  5. Substitute the point and normal gradient

    y−4=−14(x−2)y - 4 = - \frac{1}{4}(x - 2)

    Put the normal gradient and the point of contact into the straight-line formula.

  6. Simplify to the form y=mx+cy = mx + c

    y=92−x4y = \frac{9}{2} - \frac{x}{4}

    Expand and tidy up to get the normal in its neatest form. This is the equation of the normal.

Answer
y=92−x4y = \frac{9}{2} - \frac{x}{4}

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