05 6eqntangent

Tangents and Normals

. Recall, a tangent is a straight line that just touches the curve only at the point of contact.

• This point is called the point of tangency.
• At this point, the tangent line and the curve have the same gradient.

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The derivative of a function is a gradient function that gives the measure of the gradient at each point on the curve.

• If $f'(2) = 5$, then $f(x)$ has a gradient of 5 at $x = 2$

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Having the gradient, we can find the equation of the tangent for a given point on the curve.

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If $(x,\; y)$ is a point on the curve, and $f(x)$ is differentiable at $x = x_1$,

• then the equation of the tangent at $(x_1, \; y_1)$ is given by:

… … $y-y_1=f'(x_1)(x-x_1)$

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Example 1

Find the equation of the tangent of the curve:

… … $y = x^3 + \dfrac{1}{2}x^2$ at the point where $x = 1$

Solution: … … When $x = 1$

… … $y = 1^3 + \dfrac{1}{2} \times 1^2$

… … $y = 1 + \dfrac{1}{2}$

… … $y = \dfrac{3}{2}$

… … $\dfrac{dy}{dx}=3x^2+x$

… … When $x = 1$

… … $\dfrac{dy}{dx} = 3 \times 1^2 + 1$

… … $\dfrac{dy}{dx} = 4$

… … Therefore, the equation of the tangent is:

… … $y - y_1 = m(x - x_1)$

… … $y-\dfrac{3}{2}=4(x-1)$

… … $y-\dfrac{3}{2}=4x-4$

… … $y=4x-\frac{5}{2}$

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Normals

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The normal to a curve at a point is the straight line that passes through the point
and is perpendicular to the tangent at that point.
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• If m1 is the gradient of the tangent
• and m2 the gradient of the normal
• then $m_1 \times m_2 = -1$.

. Example 1 (cont)

• the normal will have gradient -\frac{1}{4}.
• (See the graph on the right)

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Therefore the equation of the normal is:

… … $y-y_1=-\dfrac{1}{f'(x_1)}(x-x_1)$

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… … $y-\dfrac{3}{2}= -\dfrac{1}{4}(x-1)$

… … $y - \dfrac{3}{2} = -\dfrac{1}{4}x + \dfrac{1}{4}$

… … $y = -\dfrac{1}{4}x + \dfrac{7}{4}$

or

… … $4y = -x + 7$

… … $4y+x=7$

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We can use the CAS calculator to find an expression for the equation of the tangent and normal. … … $x^3+\dfrac{1}{2}x^2$

• Highlight it and tap: Interactive, Calculation, tanLine.
• In the point box enter x = 1.

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Repeat the process to obtain the equation of the normal,

• choosing normal at x = 1.

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Note that the CAS calculator shows an expression instead of the equation.

• You will need to write your answer as an equation, as indicated in the answers above.

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Example 2

Find the equation of the normal to the curve with equation $y = x \sin(x)$ at the point on the curve where $x = \pi$ .

Solution:

… … When $x = \pi$

… … $y = \pi \, \sin (\pi)$

… … $y = 0$

… … $\dfrac{dy}{dx}=\sin(x)+x\cos(x)$ … … using the product rule

… … At $x=\pi, \quad \dfrac{dy}{dx}=-\pi$

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… … The gradient of the tangent is $m_1 = -\pi$

… … So, the gradient of the normal is $m_2 = \dfrac{1}{\pi}$

… … $y-0=\dfrac{1}{\pi}(x-\pi)$

… … $\pi{y}=x-\pi$

… … $x - \pi y = \pi$

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Example 3

Find the equation of the tangent to the curve with equation $y = e^{2x–4}$ that is parallel to the line $y = 5 + \frac{x}{2}$.

Solution:

… … Parallel lines have equal gradient.

… … So the tangent will have the same gradient as the line $y = 5 + \frac{x}{2}$.

… … So gradient of tangent is $m = \frac{1}{2}$.

… … Need to find the coordinates of the point where the gradient of the curve is $\frac{1}{2}$.

… … i.e. $\dfrac{dy}{dx}=\dfrac{1}{2}$

… … $\dfrac{dy}{dx}=2e^{2x-4}$ … … using the chain rule

… … $2e^{2x-4}=\dfrac{1}{2}$

… … $e^{2x-4}=\dfrac{1}{4}$

… … $2x-4=ln \left( \dfrac{1}{4} \right)$

… … $2x=4+ln \big( {4^{-1}} \big)$

… … $x=2-\dfrac{1}{2}ln(4)$

… … $x=2-ln \big( 4^{\frac{1}{2}} \big)$

… … $x=2-ln(2)$

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… … When $x=2-ln(2)$

… … $y=e^{2(2-ln(2)) - 4}$

… … $y=e^{ln(2^{-2})}$

… … $y=e^ { ln \left( \frac{1}{4} \right) }$

… … $y=\dfrac{1}{4}$

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… … Therefore the equation of the tangent is:

… … $y-\dfrac{1}{4}=\dfrac{1}{2} \Big( x- \big(2-ln(2) \big) \Big)$

… … $y -\dfrac{1}{4} = \dfrac{x}{2} -1 + \dfrac{ln(2)}{2}$

… … $y = \dfrac{x}{2} -\dfrac{3}{4} + \dfrac{ln(2)}{2}$

or

… … $4y = 2x - 3 - 2ln(2)$

… … $4y - 2x = -3 - 2ln(2)$

… … $2x - 4y = 3 + 2ln(2)$

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