Tangent line, normal line - confusing. 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!
 A: 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 $m\cdot m_{tangent} = -1$, 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.)
