6.4 Equation of a tangent to a curve | Differential calculus

6.4 Equation of a tangent to a curve (EMCH8)

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

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:

Find the derivative using the rules of differentiation.
Substitute the \(x\)-coordinate of the given point into the derivative to calculate the gradient of the tangent.
Substitute the gradient of the tangent and the coordinates of the given point into an appropriate form of the straight line equation.
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.

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

Worked example 13: Finding the equation of a tangent to a curve

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.

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

Worked example 14: Finding the equation of a tangent to a curve

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

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*}

Worked example 15: Finding the equation of a normal to a curve

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

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

Equation of a tangent to a curve

Textbook Exercise 6.5

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*}

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

\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*}

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

\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*}

parallel to the line \(y=4x-2\).

\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 \((1;1)\).

perpendicular to the line \(2y+x-4=0\).

\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 \(\left(\frac{3}{4};\frac{1}{4}\right)\).

Draw a graph of \(f\), indicating all intercepts and turning points.

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*} \(\text{Intercepts:}\\ y_{\text{int}}: x = 0, y = -3 \\ x_{\text{int}}: y=0, \\ -x^{2} +4x -3 = 0 \\ x^{2} - 4x + 3 = 0 \\ (x-3)(x-1) = 0 \\ x=3 \text{ or } x=1 \\ \text{Shape: “frown” } (a < 0) \\\)

Find the equations of the tangents to \(f\) at:

the \(y\)-intercept of \(f\).
the turning point of \(f\).
the point where \(x = \text{4,25}\).

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

Draw the three tangents above on your graph of \(f\).

Write down all observations about the three tangents to \(f\).

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.

6.3 Rules for differentiation

Table of Contents

6.5 Second derivative