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Equation of a tangent to a curve

At a given point on a curve, the gradient of the curve is equal to the gradient of the tangent to the curve.

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The derivative (or gradient function) describes the gradient of a curve at any point on the curve. Similarly, it also describes the gradient of a tangent to a curve at any point on the curve.

To determine the equation of a tangent to a curve:

  1. Find the derivative using the rules of differentiation.
  2. Substitute the \(x\)-coordinate of the given point into the derivative to calculate the gradient of the tangent.
  3. Substitute the gradient of the tangent and the coordinates of the given point into an appropriate form of the straight line equation.
  4. Make \(y\) the subject of the formula.

The normal to a curve is the line perpendicular to the tangent to the curve at a given point.

\[m_{\text{tangent}} \times m_{\text{normal}} = -1\]

Example 1: Finding the equation of a tangent to a curve

Question

Find the equation of the tangent to the curve \(y=3{x}^{2}\) at the point \(\left(1;3\right)\). Sketch the curve and the tangent.

Answer

Find the derivative

Use the rules of differentiation:

\[\begin{align} y &= 3{x}^{2} \\ & \\ \therefore \frac{dy}{dx} &= 3 \left( 2x \right) \\ &= 6x \end{align}\]

Calculate the gradient of the tangent

To determine the gradient of the tangent at the point \(\left(1;3\right)\), we substitute the \(x\)-value into the equation for the derivative.

\[\begin{align} \frac{dy}{dx} &= 6x \\ \therefore m &= 6(1) \\ &= 6 \end{align}\]

Determine the equation of the tangent

Substitute the gradient of the tangent and the coordinates of the given point into the gradient-point form of the straight line equation.

\[\begin{align} y-{y}_{1} & = m\left(x-{x}_{1}\right) \\ y-3 & = 6\left(x-1\right) \\ y & = 6x-6+3 \\ y & = 6x-3 \end{align}\]

Sketch the curve and the tangent

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Example 2: Finding the equation of a tangent to a curve

Question

Given \(g(x)= (x + 2)(2x + 1)^{2}\), determine the equation of the tangent to the curve at \(x = -1\) .

Answer

Determine the \(y\)-coordinate of the point

\[\begin{align} g(x) &= (x + 2)(2x + 1)^{2} \\ g(-1) &= (-1 + 2)[2(-1) + 1]^{2} \\ &= (1)(-1)^{2} \\ & = 1 \end{align}\]

Therefore the tangent to the curve passes through the point \((-1;1)\).

Expand and simplify the given function

\[\begin{align} g(x) &= (x + 2)(2x + 1)^{2} \\ &= (x + 2)(4x^{2} + 4x + 1) \\ &= 4x^{3} + 4x^{2} + x + 8x^{2} + 8x + 2 \\ &= 4x^{3} + 12x^{2} + 9x + 2 \end{align}\]

Find the derivative

\[\begin{align} g'(x) &= 4(3x^{2}) + 12(2x) + 9 + 0 \\ &= 12x^{2} + 24x + 9 \end{align}\]

Calculate the gradient of the tangent

Substitute \(x = -\text{1}\) into the equation for \(g'(x)\):

\[\begin{align} g'(-1) &= 12(-1)^{2} + 24(-1) + 9 \\ \therefore m &= 12 - 24 + 9 \\ &= -3 \end{align}\]

Determine the equation of the tangent

Substitute the gradient of the tangent and the coordinates of the point into the gradient-point form of the straight line equation.

\[\begin{align} y-{y}_{1} & = m\left(x-{x}_{1}\right) \\ y-1 & = -3\left(x-(-1)\right) \\ y & = -3x - 3 + 1 \\ y & = -3x - 2 \end{align}\]

Example 3: Finding the equation of a normal to a curve

Question

  1. Determine the equation of the normal to the curve \(xy = -4\) at \(\left(-1;4\right)\).
  2. Draw a rough sketch.

Answer

Find the derivative

Make \(y\) the subject of the formula and differentiate with respect to \(x\):

\[\begin{align} y &= -\frac{4}{x} \\ &= -4x^{-1} \\ & \\ \therefore \frac{dy}{dx} &= -4 \left( -1x^{-2} \right) \\ &= 4x^{-2} \\ &= \frac{4}{x^{2}} \end{align}\]

Calculate the gradient of the normal at \(\left(-1;4\right)\)

First determine the gradient of the tangent at the given point:

\[\begin{align} \frac{dy}{dx} &= \frac{4}{(-1)^{2}} \\ \therefore m &= 4 \end{align}\]

Use the gradient of the tangent to calculate the gradient of the normal:

\[\begin{align} m_{\text{tangent}} \times m_{\text{normal}} &= -1 \\ 4 \times m_{\text{normal}} &= -1 \\ \therefore m_{\text{normal}} &= -\frac{1}{4} \end{align}\]

Find the equation of the normal

Substitute the gradient of the normal and the coordinates of the given point into the gradient-point form of the straight line equation.

\[\begin{align} y-{y}_{1} & = m\left(x-{x}_{1}\right) \\ y-4 & = -\frac{1}{4}\left(x-(-1)\right) \\ y & = -\frac{1}{4}x - \frac{1}{4} + 4\\ y & = -\frac{1}{4}x + \frac{15}{4} \end{align}\]

Draw a rough sketch

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Exercise 1: Equation of a tangent to a curve

Determine the equation of the tangent to the curve defined by \(F(x)=x^{3}+2x^{2}-7x+1\) at \(x=2\).

\[\begin{align} \text{Gradient of tangent }&= F'(x) \\ F'(x) &=3x^{2} +4x - 7 \\ F'(2) &=3(2)^{2} + (4)(2) -7 \\ &=13 \\ \therefore \text{Tangent: } y &=13x +c \end{align}\]

where \(c\) is the \(y\)-intercept.

Tangent meets \(F(x)\) at \((2;F(2))\)

\[\begin{align} F(2) &=(2)^{3} + 2(2)^{2} - 7(2) +1 \\ &= 8 + 8 -14 +1 \\ &=3 \\ \text{Tangent: } 3 &=13(2) + c \\ \therefore c &= - 23 \\ y & = 13x - 23 \end{align}\]

Determine the point where the gradient of the tangent to the curve:

  1. \(f(x)=1-3x^{2}\) is equal to 5.

  2. \(g(x)=\frac{1}{3}x^{2}+2x+1\) is equal to 0.

  1. \[\begin{align} \text{Gradient of tangent } = f'(x) = -6x \\ \therefore -6x &= 5 \\ \therefore x &= - \frac{5}{6} \\ \text{And } f\left(- \frac{5}{6} \right) &=1-3 \left( - \frac{5}{6} \right)^{2} \\ &=1-3 \left( \frac{25}{36} \right) \\ &=1 - \frac{25}{12} \\ &= - \frac{13}{12} \\ \therefore & \left( - \frac{5}{6};- \frac{13}{12} \right) \end{align}\]

  2. \[\begin{align} \text{Gradient of tangent } = g'(x) = \frac{2}{3}x+2 \\ \therefore \frac{2}{3}x+2 &=0 \\ \frac{2}{3}x &= -2\\ \therefore x&=-2 \times \frac{3}{2} \\ &=-3 \\ \text{And } g(-3) &= \frac{1}{3}(-3)^{2}+2(-3)+1 \\ &= \frac{1}{3}(9)-6+1 \\ &= 3-6+1 \\ &= -2 \\ \therefore & (-3;-2) \end{align}\]

Determine the point(s) on the curve \(f(x)=(2x-1)^{2}\) where the tangent is:

  1. parallel to the line \(y=4x-2\).
  2. perpendicular to the line \(2y+x-4=0\).
  1. \[\begin{align} \text{Gradient of tangent }&= f'(x) \\ f(x)&=(2x-1)^{2} \\ &= 4x^{2}-4x+1 \\ \therefore f'(x)&= 8x-4 \\ \text{Tangent is parallel to } y&=4x-2 \\ \therefore m&=4 \\ \therefore f'(x) = 8x-4 &= 4 \\ 8x &= 8 \\ x & = 1\\ \text{For } x=1: \quad y & = (2(1)-1)^{2} \\ & = 1 \end{align}\]

    Therefore, the tangent is parallel to the given line at the point \((1;1)\).

  2. \[\begin{align} \text{Perpendicular to } 2y + x - 4 &= 0 \\ y&= -\frac{1}{2}x+2\\ \therefore \text{ gradient of } \perp \text{ line } & = 2 \quad (m_1 \times m_2 = -1) \\ \therefore f'(x) &= 8x-4 \\ \therefore 8x-4 &=2\\ 8x&=6\\ x&=\frac{3}{4} \\ \therefore y&=\left[2\left(\frac{3}{4}\right)-1\right]^{2} \\ &=\frac{1}{4} \\ \therefore \left(\frac{3}{4};\frac{1}{4}\right) \end{align}\]

    Therefore, the tangent is perpendicular to the given line at the point \(\left(\frac{3}{4};\frac{1}{4}\right)\).

Given the function \(f\): \(y=-x^{2}+4x-3\).

  1. Draw a graph of \(f\), indicating all intercepts and turning points.
  2. Find the equations of the tangents to \(f\) at:
    1. the \(y\)-intercept of \(f\).
    2. the turning point of \(f\).
    3. the point where \(x = \text{4,25}\).
  3. Draw the three tangents above on your graph of \(f\).
  4. Write down all observations about the three tangents to \(f\).
  1. Complete the square:

    \[\begin{align} y&=-[x^{2}-4x+3] \\ &=-[(x-2)^{2}-4+3] \\ &=-(x-2)^{2}+1\\ \text{Turning point}:&(2;1) \end{align}\]

    \(\text{Intercepts:}\\ y_{\text{int}}: x = 0, y = -3 \\ x_{\text{int}}: y=0, \\ -x^{2} +4x -3 = 0 \\ x^{2} - 4x + 3 = 0 \\ (x-3)(x-1) = 0 \\ x=3 \text{ or } x=1 \\ \text{Shape: “frown” } (a < 0) \\\)
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    1. \[\begin{align} y_{\text{int}}: (0;-3) \\ m_{\text{tangent}} = f'(x) &= -2x + 4 \\ f'(0) &=-2(0) + 4 \\ \therefore m &=4\\ \text{Tangent }y&=4x+c\\ \text{Through }(0;-3) \therefore y&=4x-3 \end{align}\]

    2. \[\begin{align} \text{Turning point: } (2;1) \\ m_{\text{tangent}} = f'(2) &= -2(2) + 4 \\ &=0\\ \text{Tangent equation } y &= 1 \end{align}\]

    3. \[\begin{align*} \text{If } x &=\text{4,25} \\ f(\text{4,25})&=-\text{4,25}^{2}+4(\text{4,25})-3 \\ &= -\text{4,0625} \\ m_{\text{tangent}} \text{ at } x&= \text{4,25} \\ m&=-2(\text{4,25})+4\\ &=-\text{4,5} \\ \text{Tangent }y&=-\text{4,5}x+c\\ \text{Through }(\text{4,25};-\text{4,0625}) \\ -\text{4,0625}&=-\text{4,5}(\text{4,25})+c\\ \therefore c&= \text{15,0625} \\ y&=-\text{4,5}x+\text{15,0625} \end{align*}\]

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  3. Tangent at \(y_{\text{int}}\) (blue line): gradient is positive, the function is increasing at this point.

    Tangent at turning point (green line): gradient is zero, tangent is a horizontal line, parallel to \(x\)-axis.

    Tangent at \(x=\text{4,25}\) (purple line): gradient is negative, the function is decreasing at this point.