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I am practicing finding tangent and normal line. The tangent/normal line is usually to some graph, and parallel/perpendicular to some other line at the same time. Not that complicated.

Can someone, please, verify my solutions? The more I study the less I know, and I don't even know how much I don't know. Thank you!

But I'm confused, mostly about the $a$ "directional coefficient" of $y=ax+b$ part.

Let's say the task is: Find tangent and normal lines to the function $f(x) = ln(x+1)$, parallel&perpendicular to $y=frac{1}{2}x$. (so 4 variations in total.)

What I know: I do know that if a line has to be parallel to other line, then the $a$ coefficients have to be the same. And if perpendicular, then I have to inverse it and change the sign.

What I am not sure of: how to properly write the equations. How to properly substitute into the equations. I am doing mistakes here related to $a$ coefficient.

I am looking for an algorithmic way so I can easily understand and remember how to solve it.

$f(x) = ln(x+1)$

$f'(x) = frac{1}{x+1}$

1. Finding tangent parallel to $y=frac{1}{2}x$.

$frac{1}{x+1} = frac{1}{2}$

$x = 1 = x_{0}$

$f(x_{0}) = f(1) = ln2$

$Rightarrow y = frac{1}{2}x - frac{1}{2} + ln2$

2. Finding tangent perpendicular to $y=frac{1}{2}x$

$frac{1}{x+1} = -2$

$x = 1 = x_{0}$

$f(x_{0}) = f(1) = ln2$

$Rightarrow y - ln2 = -2(x-x_{0})$

$Rightarrow y = -2x + 2 + ln2$

3. Finding normal line parallel to $y = frac{1}{2}x$

$frac{1}{x+1} = -2$

$x = frac{-3}{2} = x_{0}$

$f(x_{0}) = f(-frac{3}{2})= ln(frac{-1}{2})$ - does not exist

$y = frac{1}{2}(x-x_{0})$

$y = frac{1}{2}x + frac{3}{4}$

4. Finding normal line perpendicular to $y = frac{1}{2}x$

$frac{1}{x+1} = -2$

$x = frac{-3}{2} = x_{0}$

$f(x_{0}) = f(frac{-3}{2}) = ln(frac{-1}{2})$ - does not exist

$y - 0 = -2(x-x_{0})$

$y = -2x - 3$

Thanks for your time!

It’s not that difficult once you get the hang of the plan. Basically, in a question involving a function $f(x)$ and finding its tangent/normal lines at a particular point, you’d want to carry out the following steps:

• Find the general formula for a slope at a given point by computing the derivative $f’(x)$.


• Find the $x$-coordinate that satisfies the given conditions. Then, plug in the $x$-value obtained back into $f(x)$ to get the $y$-coordinate.


• Use point-slope form, or $y-y_1 = m(x-x_1)$ to find the equation.

which you seem to know.

For the example you’ve given, we have $f(x) = ln(x+1)$, so $f’(x) = frac{1}{x+1}$ as you’ve found.

$(a)$ For the tangent parallel to $y = frac{1}{2}x$, you immediately obtain $m = frac{1}{2}$, so $m_{tangent} = frac{1}{2}$ as well.

• $$f’(x) = frac{1}{2} iff frac{1}{2} = frac{1}{x+1} iff x = 1$$




• $$f(1) = ln(1+1) = ln 2$$




• $$y-y_1 = m(x-x_1) implies y-ln 2 = frac{1}{2}(x-1) implies boxed{y = frac{1}{2}x-frac{1}{2}+ln 2}$$

$(b)$ For the tangent perpendicular to $y = frac{1}{2}x$, $m = frac{1}{2}$, but $mcdot m_{tangent} = -1$, so $m_{tangent} = -2$.

• $$f’(x) = -2 iff -2 = frac{1}{x+1} iff x = -frac{3}{2} = -1.5$$


• $$f(-1.5) = ln(1+(-1.5)) = ln(-0.5)$$

But wait, $ln(-0.5)$ is not defined. Clearly there is an issue.

The issue is that $f’(x)$ can’t possibly be negative. Recall that in logs, the argument must be positive:

$$f(x) = ln(1+x); quad x > -1$$

$$f’(x) = frac{1}{1+x}; quad x > -1$$

So $f’(x)$ can’t be negative, which is why we have this problem. Therefore, you can’t have a tangent with a negative slope for this function. You can observe this in the graph of $f(x) = ln(x+1)$, which never curves downward at any point.

You use the exact same plan for normal lines, except you also have to note that

$$m_{tangent}cdot m_{normal} = -1 iff m_{normal} = -frac{1}{m_{tangent}}$$

Since for the given problem, the tangent line’s slope is always positive, the normal line’s slope will always be negative.

Another tip I can give you is to use the connection between tangent lines and normal lines to your advantage and save time.

For instance, you can do the following:

• Find $x$ such that the normal line is perpendicular to $y = frac{1}{2}x$.


• 
  $implies$ Find $x$ such that the tangent line is parallel to $y = frac{1}{2} x$.

and

• Find $x$ such that the normal line is parallel to $y = frac{1}{2}x$.


• 
  $implies$ Find $x$ such that the tangent line is perpendicular to $y = frac{1}{2} x$.

Which essentially skips the entire first step since you already calculated those points in the first two parts. (Of course, this is only for finding the $x$-coordinate. Don’t confuse this for the slopes and equations of the lines themselves! They’re completely different lines, but intersect at the same point.)