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

Please help. Thanks.

Printable View

- Apr 20th 2013, 04:27 AMonehundredpercenteffortNatural 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.

Please help. Thanks. - Apr 20th 2013, 08:11 AMemakarovRe: Natural Logarithm: find the point on the curve
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 AMPlatoRe: Natural Logarithm: find the point on the curve
- Apr 20th 2013, 10:52 AMhollywoodRe: 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 AMonehundredpercenteffortRe: 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 AMPlatoRe: Natural Logarithm: find the point on the curve
- Apr 21st 2013, 07:28 AMhollywoodRe: 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