# Natural Logarithm: find the point on the curve

• Apr 20th 2013, 04:27 AM
onehundredpercenteffort
Natural Logarithm: find the point on the curve
Question: Find the point on the curve y=e^2x+1 where the tangent is parallel to the line y=2ex.

• Apr 20th 2013, 08:11 AM
emakarov
Re: Natural Logarithm: find the point on the curve
Quote:

Originally Posted by onehundredpercenteffort
Question: Find the point on the curve y=e^2x+1 where the tangent is parallel to the line y=2ex.

The graph of $\displaystyle y = e^2x+1$ is a straight line, so a tangent to it is itself. Since the slope of $\displaystyle y = e^2x+1$, which is $\displaystyle e^2$, is different from the slope of $\displaystyle y = 2ex$, which is $\displaystyle 2e$, no tangent to $\displaystyle y = e^2x+1$ is parallel to $\displaystyle y = 2ex$.
• Apr 20th 2013, 08:18 AM
Plato
Re: Natural Logarithm: find the point on the curve
Quote:

Originally Posted by onehundredpercenteffort
Question: Find the point on the curve y=e^2x+1 where the tangent is parallel to the line y=2ex.

Is the function $\displaystyle e^{2x}+1\text{ or }e^{2x+1}$ or something else altogether?
• Apr 20th 2013, 10:52 AM
hollywood
Re: Natural Logarithm: find the point on the curve
Whatever the function, you need to take the derivative and set it equal to 2e.

- Hollywood
• Apr 21st 2013, 06:06 AM
onehundredpercenteffort
Re: Natural Logarithm: find the point on the curve
i understand that i have to find the derivative which was y'=2e^2x+1 and since the tangent is parallel, the slopes will be the same. But how can i find the point??
• Apr 21st 2013, 06:23 AM
Plato
Re: Natural Logarithm: find the point on the curve
Quote:

Originally Posted by onehundredpercenteffort
i understand that i have to find the derivative which was y'=2e^2x+1 and since the tangent is parallel, the slopes will be the same. But how can i find the point??

Why do you refuse to tell us what the function really is?

What is the correct way to read the function? See reply #3.
• Apr 21st 2013, 07:28 AM
hollywood
Re: Natural Logarithm: find the point on the curve
You set y' = 2e^(2x+1) = 2e and solve for x. The slope of the other curve, y=2ex, is 2e. So x=0, right?

All three of us were confused about what the function was; you really should have cleared that up in addition to asking for clarification.

- Hollywood