Question

Determine the point on the graph of y = In 2x at which the tangent line is perpendicular to

Answer 1

Answer:

$(\frac{1}{4},-\ln(2))$

Step-by-step explanation:

Let's first differentiate $y=\ln(2x)$.

This gives us $y'=\frac{(2x)'}{2x}=\frac{2}{2x}=\frac{1}{x}$.  This gives us the slope of any tangent line to any point on the curve of $y=\ln(2x)$.

Let $(a,b)$ be a point on $y=\ln(2x)$ such that the tangent line at that point is perpendicular to $x+4y=1$.

Let's find the slope of this perpendicular line so we can determine the slope of the tangent line. Keep in mind, that perpendicular lines (if not horizontal to vertical lines or vice versa) have opposite reciprocal slopes.

Let's begin.

$x+4y=1$

Subtract $x$ on both sides:

$4y=-x+1$

Divide both sides by 4:

$y=\frac{-x}{4}+\frac{1}{4}$

The slope is -1/4.

This means the line perpendicular to it, the slope of the line we wish to find, is 4.

So we want the following to be true:

$\frac{1}{x} \text{ at } x=a$ to be $4$.

So we are going to solve the following equation:

$\frac{1}{a}=4$

Multiply both sides by $a$:

$1=4a$

Divide both sides by 4:

$\frac{1}{4}=a$

So now let's find the corresponding $y$-coordinate that I called $b$ earlier for our particular point that we wished to find.

$y=\ln(2x)$ for $x=a=\frac{1}{4}$:

$y=\ln(2\cdot \frac{1}{4})$ (this is our $b$)

$y=\ln(\frac{1}{2})$

$y=\ln(1)-\ln(2)$

$y=0-\ln(2)$

$y=-\ln(2)$

So the point that we wished to find is $(\frac{1}{4},-\ln(2))$.

---------------------Verify--------------------------------

What is the line perpendicular to the tangent line to the curve $y=\ln(2x)$ at $(\frac{1}{4},-\ln(2))$?

Let's find the slope formula for our tangent lines to this curve:

$y'=\frac{1}{x}$

$y'=\frac{1}{x}$ evaluated at $x=\frac{1}{4}$:

$y'=\frac{1}{\frac{1}{4}}=4$

This says the slope of this tangent line is 4.

A line perpendicular this will have slope -1/4.

So we know our line will be of the form:

$y=\frac{-1}{4}x+c$

Multiply both sides by 4:

$4y=-1x+4c$

Add $1x$ on both sides:

$4y+1x=4c$

Reorder using commutative property:

$1x+4y=4c$

Use multiplicative identity property:

$x+4y=4c$

As we see the line is in this form. We didn't need to know about the $y$-intercept,$c$, of this equation.

Answer 2

Answer:

  (1/4, ln(1/2))

Step-by-step explanation:

The slope of the given line is -1/4, so the perpendicular line will have a slope of -1/(-1/4) = 4.

The slope of the given function is its derivative:

  y' = 2/(2x) = 1/x

That will have a value of 4 when x = 1/4.

The point on the graph where the slope is 4 is (x, y) = (1/4, ln(1/2)).