Equation for the Normal to a Curve – Examples with answers

We can find the equation of the normal line using the general form $latex y=mx+b$, where m is the slope and b is the y-intercept.

Now, the slope of the normal is equal to $latex -\frac{1}{m_{1}}$, where $latex m_{1}$ is the slope of the tangent line. In turn, the slope of the tangent line is equal to $latex f'(1)$ (we evaluate the derivative at the given point). Then, we have:

$latex f(x)=x^2$

$latex f'(x)=2x$

$latex m_{1}=f'(1)=2(1)$

$latex m_{1}=2$

∴ $latex m=-\frac{1}{2}$

This means we have the equation $latex y=-\frac{1}{2}x+b$. Therefore, we use the point (1, 2) in the equation to determine the value of b:

$latex y=-\frac{1}{2}x+b$

$latex 2=-\frac{1}{2}(1)+b$

$latex b=\frac{5}{2}$

The equation of the normal line at the point (1, 2) is $latex y=-\frac{1}{2}x+\frac{5}{2}$.

To find the equation of the normal line, we have to start by finding the derivative of the function. Thus, we have:

$latex f(x)=x^3-6x$

$latex f'(x)=3x^2-6$

Now, we can use the derivative to find the slope of the tangent line at the point (-1, 1) by evaluating $latex f'(-1)$:

$latex m_{1}=f'(-1)=3(-1)^2-6$

$latex m_{1}=-3$

The slope of the normal line is equal to $latex m=\frac{1}{3}$. Therefore, we have the equation $latex y=\frac{1}{3}x+b$. Now, we find the value of b, using the point (-1, 1) in the equation:

$latex y=\frac{1}{3}x+b$

$latex 1=\frac{1}{3}(-1)+b$

$latex b=\frac{4}{3}$

The equation of the normal line at the point (-1, 1) is $latex y=\frac{1}{3}x-\frac{4}{3}$.

We start by finding the slope of the tangent line. For this, we obtain the derivative of the function:

$latex f(x)=2x^3-7x^2$

$latex f'(x)=6x^2-14x$

The slope of the tangent line at the point (2, 3) is given by $latex f'(2)$:

$latex m_{1}=f'(2)=6(2)^2-14(2)$

$latex m_{1}=24-28$

$latex m_{1}=-4$

The slope of the normal line is $latex m=\frac{1}{4}$. Then, we have the equation $latex y=\frac{1}{4}x+b$. Now, we use this equation with the point (2, -3) to find the value of b:

$latex y=\frac{1}{4}x+b$

$latex -3=\frac{1}{4}(2)+b$

$latex b=-\frac{7}{2}$

The equation of the normal line at the point (2, -3) is $latex y=\frac{1}{4}x-\frac{7}{2}$.

The derivative of the function is:

$latex f(x)=x^3+6x^{-1}$

$latex f'(x)=3x^2-6x^{-2}$

$latex f'(x)=3x^2-\frac{6}{x^2}$

Now, we use the derivative to find the slope of the tangent at the point (1, -2). This is equal to $latex f'(1)$:

$latex m_{1}=f'(1)=3(1)^2-\frac{6}{1^2}$

$latex =3-6$

$latex m_{1}=-3$

The slope of the normal line is $latex m=\frac{1}{3}$. Therefore, we have the equation $latex y=\frac{1}{3}x+b$. Then, we use the point (1, -2) to find the value of b:

$latex y=\frac{1}{3}x+b$

$latex -2=\frac{1}{3}(1)+b$

$latex b=-\frac{7}{3}$

The equation of the normal line at the point (1, -2) is $latex y=\frac{1}{3}x-\frac{7}{3}$.

We have to start by finding the derivative of the given function:

$latex f(x)=-x^{-2}+x^{\frac{1}{2}}$

$$f'(x)=2x^{-3}+\frac{1}{2}x^{-\frac{1}{2}}$$

$$f'(x)=\frac{2}{x^3}+\frac{1}{2\sqrt{x}}$$

Now, we use the derivative to find the slope of the tangent at the point (1, -2). This is equal to $latex f'(1)$:

$$m_{1}=f'(1)=\frac{2}{(1)^3}+\frac{1}{2\sqrt{1}}$$

$latex =2+\frac{1}{2}$

$latex m_{1}=\frac{5}{2}$

The slope of the normal line is $latex m=-\frac{2}{5}$. Then, we have the equation $latex y=-\frac{2}{5}x+b$, so we use the point (1, -2) to find the value of b:

$$y=-\frac{2}{5}x+b$$

$$-2=-\frac{2}{5}(1)+b$$

$latex b=-\frac{8}{5}$

The equation of the normal line at the point (1, -2) is $latex y=-\frac{2}{5}x-\frac{8}{5}$.

The derivative of the function is:

$latex f(x)=\sin(x)-\cos(x)$

$latex f'(x)=\cos(x)+\sin(x)$

The slope of the tangent line at the point (0, 1) is equal to $latex f'(0)$. Then, we have:

$latex m_{1}=f'(0)=\cos(0)+\sin(0)$

$latex m_{1}=1+0$

$latex m_{1}=1$

The slope of the normal line is equal to $latex m=-1$. Therefore, we have the equation $latex y=-x+b$ and now we use this equation with the point (0, 1) to find the value of b:

$latex y=-x+b$

$latex 1=-0+b$

$latex b=1$

The equation of the normal line at the point (0, 1) is $latex y=-x+1$.

The derivative of the function is:

$latex f(x)=-2\sin(2x)+\cos(3x)$

$latex f'(x)=-4\cos(2x)-3\sin(3x)$

Using the derivative, we can find the slope of the tangent line at (0, 1) by evaluating $latex f'(0)$:

$latex m_{1}=f'(0)=-4\cos(2(0))-3\sin(3(0))$

$latex m_{1}=-4+0$

$latex m_{1}=-4$

The slope of the normal line is $latex m=\frac{1}{4}$. Therefore, we have the equation $latex y=\frac{1}{4}x+b$. Now, we use this equation with the point (0, 1) to determine the value of b:

$latex y=\frac{1}{4}x+b$

$latex 1=0+b$

$latex b=1$

The equation of the normal line at the point (0, 1) is $latex y=\frac{1}{4}x+1$.

We start by finding the derivative of the function:

$latex f(x)=x^2-3x+1$

$latex f'(x)=2x-3$

We don’t know the coordinates of the point directly, but we do know that the line is normal to the point where the curve intersects the y-axis. This happens when the coordinates in x are equal to 0. Thus, we have:

$latex y=x^2-3x+1$

$latex y=0^2-3(0)+1$

$latex y=1$

Therefore, the point is (0, 1). Now, we use the derivative of the function to evaluate $latex f'(0)$ and find the slope of the tangent:

$latex m_{1}=f'(0)=2(0)-3$

$latex m_{1}=-3$

The slope of the normal line is $latex m=\frac{1}{3}$. Therefore, we have the equation $latex y=\frac{1}{3}x+b$. Using the point (0, 1) we find the value of b:

$latex y=\frac{1}{3}x+b$

$latex 1=\frac{1}{3}(0)+b$

$latex b=1$

The equation of the normal line at the point (0, 1) is $latex y=\frac{1}{3}x+1$.

We have to start by finding the points where the function intersects the x-axis. This happens when the value of y is 0. Therefore, we have:

$latex x^2-5x+4=0$

$latex (x-4)(x-1)=0$

The function cuts the x-axis when $latex x=4$ and $latex x=1$. This means that we have the points (4, 0) and (1, 0).

Now, we use the derivative of the function to find the slopes of the tangent lines by evaluating $latex f'(4)$ and $latex f'(1)$:

$latex f(x)=x^2-5x+4$

$latex f'(x)=2x-5$

$latex f'(4)=2(4)-5=3$

$latex f'(1)=2(1)-5=-3$

The slopes of the normal lines are $latex m_{1}=-\frac{1}{3}$ and $latex m_{2}=\frac{1}{3}$. Thus, we use the form $latex y=mx+b$ with the slopes found and the coordinates of each point to determine the values of b:

$latex 0=\frac{1}{3}(4)+b_{1}$

$latex b_{1}=-\frac{4}{3}$

$latex 0=-\frac{1}{3}(1)+b_{2}$

$latex b_{2}=\frac{1}{3}$

The equations of the normal lines are $latex y=-\frac{1}{3}x-\frac{4}{3}$ and $latex y=\frac{1}{3}x+\frac{ 1}{3}$.

We have that the lines are normal at the points where $latex y=9$. This means that we have to find the values of x when $latex y=9$:

$latex x^2=9$

$latex x=\pm \sqrt{9}$

$latex x=3~~$ or $latex ~~x=-3$

Now that we know the coordinates of the points, we use the derivative of the function to find the slopes of the tangent lines by evaluating $latex f'(3)$ and $latex f'(-3)$.

$latex f(x)=x^2$

$latex f'(x)=2x$

$latex f'(3)=2(3)=6$

$latex f'(-3)=2(-3)=-6$

The slopes of the normal lines are $latex m_{1}=-\frac{1}{6}$ and $latex m_{2}=\frac{1}{6}$.

Now, we use the form $latex y=mx+b$ with the slopes found and the coordinates of each point (the y values are 9 in both cases) to find the values of b:

$latex 9=-\frac{1}{6}(3)+b_{1}$

$latex b_{1}=\frac{19}{2}$

$latex 9=\frac{1}{6}(-3)+b_{2}$

$latex b_{2}=\frac{19}{2}$

The equations of the normal lines are $latex y=-\frac{1}{6}x+\frac{19}{2}$ and $latex y=\frac{1}{6}x+\frac{19 }{2}$.