Find the equation of the normal line to the curve $y = \sqrt{1+4x}$ at $x=2$.

Question

Find the equation of the normal line to the curve $y = \sqrt{1+4x}$ at $x=2$.

So, to begin we get rid of the square root right?

$y=(1+4x)^{-1/2}$

Then, the power rule?

$y’= -1/2(1+4x)$

Then, don’t we multiply by the derivative of the inside of the parenthesis?

$y’ = -1/2 (1+4x) \cdot (4)$

$y’ = -2 (1+4x)$

$y’ = -2 - 8x$

Then, we set our derivative to 0 to find the slope.

$0 = -2 - 8x$

$-1/4 = x$ This is our slope!

Then, the normal means it’s the reciprocal of our slope, so it’s actually $-4$.

Then, we plug our $x$ value into our original equation to get our $y$ value.

$y= \sqrt{1+4(1/4)}$

$y= \sqrt{2}$ (our teacher said we aren’t allowed to have decimals in our answers during this section)

Then we plug everything in.

$\sqrt{2} = -4 \cdot 2 + b$

$b= \sqrt{2} + 8$

Is this correct?

Answer 1

Find the equation of the normal line using point and slope.

The point on the curve where the normal goes through is $(2,3)$

To get the slope of the normal by taking the negative reciprocal of the slope of the tangent, which you find by differentiating.

$$y=(1+4x)^\frac12 \implies y' = \frac 12 (1+4x)^{-\frac 12}(4)$$

at $x=2$ the slope of the tangent will be $\frac 23$ so the slope of the normal line will be $-\frac 32 $

Answer 2

There seems to be a few gaps in your reasoning. I don't think it's appropriate to address all of them in this reply, but I hope the following solution will help.

Let's try to work this one out from scratch.

Suppose I asked you to find out what the equation of a line is, given a point $(x_1,y_1)$ on the line, and the gradient $k$. Would you be able to do it?

Well this should be straightforward right? You know that the equation of the line should be $y = mx + c$, where m is the gradient, and $c$ is the y intercept. So all you need to do is to find out what $c$ is. Given that you have $(x_1,y_1)$ and the gradient $k$, you have:

$$ y_1 = kx_1 + c \implies c = y_1 - kx_1 $$

Then you'd be able to figure out the equation of the line, given that all the information you need for the line is its gradient and y-intercept, both of which you now know.

So, going back to your original problem, you can easily find out a point on your normal line by setting $$ \begin{align} x_1 &= 2 \\ y_1 &= \sqrt{1+4x_1} = 3 \end{align} $$ Now, all you need to do is to find out what the gradient of the normal is in order to find out the equation of the normal (since we have shown above that given any point on the line and the gradient, that we'd be able to find the y-intercept)

To find out what the gradient of the normal is, we will first find out what the gradient of the tangent to the curve $y = \sqrt{1+4x}$ is at $x=2$. Using the usual rules of differentiation, we obtain: $$ \begin{align} \frac{d y}{d x} = \frac{2}{\sqrt{1+4x}} \end{align} $$ And since we want this quantity at x = 2, we have: $$ \begin{align} \frac{d y}{d x}\Big|_{x = 2} &= \frac{2}{\sqrt{1+4\times2}}\\ &= \frac{2}{3} \end{align} $$ Now, using the fact that the slope $m_2$ of a line which is perpendicular to another line with slope $m_1$ is given by: $$ m_2 = -\frac{1}{m_1} $$ Thus, we know that the slope $k_1$ of the normal, which is perpendicular (by definition) to the tangent of $y = \sqrt{1+4x}$ at $x = 2$ (which has slope $\frac{2}{3}$), has to be given by: $$ k_1 = -\frac{1}{2/3} = -\frac{3}{2} $$ Now that we have a point $(2,\sqrt{5})$ on our line, as well as the gradient ($-\frac{3}{2}$) of the line, we can find out the y-intercept $c_1$ of the line, which is given by: $$ c_1 = 3 + 2\frac{3}{2} = 3 + 3 = 6 $$ Thus, the line can be written as: $$ y = -\frac{3}{2}x + 6 $$

I do think that you should revisit some of the concepts in differentiation to get a clearer (geometrical) understanding of what is happening here. It can help you avoid errors (such as setting the derivative to zero, which is a technique used for finding stationary points, not to calculate gradients).

Answer 3

$x=2\to y=3$

$$y=\sqrt{1+4x} \to x= \frac 14y^2-\frac 14$$ $$\to \frac{dx}{dy}=\frac y2$$

The gradient of the normal at a point is $-\frac{dx}{dy}$ at that point. Then you just need to find the correct constant.

Answer 4

The equation of the normal line to $y=f(x)$ at the point $x_0$ is: $$y=f(x_0)-\frac1{f'(x_0)}(x-x_0)$$

You want to find the normal line to $y=\sqrt{1+4x}$ at the point $x_0=2$. Applying the formula: $$f'(x)=(\sqrt{1+4x})'=\frac{2}{\sqrt{1+4x}} \Rightarrow f'(2)=\frac23;\\ y=f(2)-\frac1{f'(2)}(x-2)=3-\frac32(x-2)=-\frac32x+6.$$

The equation of the tangent line to $y=f(x)$ at the point $x_0$ is: $$y=f(x_0)+f'(x_0)(x-x_0)$$

If you want to find the tangent line to $y=\sqrt{1+4x}$ at the point $x_0=2$. Applying the formula: $$f'(x)=(\sqrt{1+4x})'=\frac{2}{\sqrt{1+4x}} \Rightarrow f'(2)=\frac23;\\ y=f(2)+f'(2)(x-2)=3+\frac23(x-2)=\frac23x+\frac53.$$

See the graph:

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