3 The tangent line to a curve

Let the equation of the tangent line to the curve $y=f\left(x\right)$ at the point $\left(x_0,y_0\right)$ be:

$y=mx+c$

where $m$ and $c$ are constants to be found. The line just touches the curve $y=f\left(x\right)$ at the point $\left(x_0,y_0\right)$ so, at this point both must have the same value for the derivative. That is:

$m=\frac{df}{dx}{\left|\right.}_{x_0}$

Since we know (in any particular case) $f\left(x\right)$ and the value $x_0$ we can readily calculate the value for $m$ . The value of $c$ is found by using the fact that the tangent line and the curve pass through the same point $\left(x_0,y_0\right)$ .

$y_0=m{x}_0+c\text{and}{y}_0=f\left(x_0\right)$

Thus $m{x}_0+c=f\left(x_0\right)\text{leading to}c=f\left(x_0\right)-m{x}_0$

Example 1

Find the equation of the tangent line to the curve $y=x^2$ at the point (1,1).

Solution

Method 1

Here $f\left(x\right)=x^2$ and $x_0=1$  thus  $\frac{df}{dx}=2x\therefore m=\frac{df}{dx}{\left|\right.}_{x_0}=2$

Also $c=f\left(x_0\right)-m{x}_0=f\left(1\right)-m=1-2=-1$ . The tangent line has equation $y=2x-1.$

Method 2

$y_0=f\left(x_0\right)=f\left(1\right)=1^2=1$

The tangent line has equation $y-1=2\left(x-1\right)\to y=2x-1$

Task!

Find the equation of the tangent line to the curve $y={\text{e}}^x$ at the point $x=0$ . The curve and the line are displayed in the following figure:

First specify $x_0$ and $f$ :

Answer

Now obtain the values of $\frac{df}{dx}{\left|\right.}_{x_0}$  and  $f\left(x_0\right)-m{x}_0$ :

Answer

Now obtain the equation of the tangent line:

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Task!

Find the equation of the tangent line to the curve $y=sin3x$ at the point $x=\frac{\pi }{4}$ and find where the tangent line intersects the $x$ -axis. See the following figure:

First specify $x_0$ and $f$ :

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Now obtain the values of $\frac{df}{dx}{\left|\right.}_{x_0}$  and $f\left(x_0\right)-m{x}_0$ correct to 2 d.p.:

Answer

Now obtain the equation of the tangent line:

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Where does the line intersect the $x$ -axis?

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