Solving for Parallel Tangent Lines: A Confusing Example

[zenity]

I'm kinda stuck on this question... my text gives me a simple example, but it's far from enlightening. Just wondered if I could get some help!

"At what point on the curve y = 1 + 2e^x - 3x is the tangent line parallel to the line 3x -y =5?"

Would I just need to simply compare the slopes?

I found the derivative of the curve: y'= 2e^x - 3

So I suppose you just set 2e^x = 3... then you get x = ln(3)/2 ?

Thanks~

 

[Phymath]

parallel means same slope, so u found $$\frac{dy}{dx} = 2e^x - 3$$
$$y=-5+3x, \frac{dy}{dx} = 3$$
$$3 = 2e^x - 3$$
$$3 = e^x$$
$$ln(3) = x$$

 

[zenity]

Oh, so I had to compute the derivatives for both slopes, then compare them? Interesting... thanks! I have to digest this info now.

 

[xanthym]

I'm kinda stuck on this question... my text gives me a simple example, but it's far from enlightening. Just wondered if I could get some help!

"At what point on the curve y = 1 + 2e^x - 3x is the tangent line parallel to the line 3x -y =5?"

Would I just need to simply compare the slopes?

I found the derivative of the curve: y'= 2e^x - 3

So I suppose you just set 2e^x = 3... then you get x = ln(3)/2 ?
-------> Should be -----> 2e^x - 3 = 3 -----------> x = ln(3)
Thanks~

zenity --

Your original approach was CORRECT. You just made the careless error shown above. Method is to determine slope of the line (compute dy/dx of the line OR determine by inspection (like you did) from the line's {y = mx + b} equation that the slope is "3") and then to equate this value to the curve's tangent slope (found by differentiating the curve's equation like you did).

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[zenity]

so the point would be... ( ln(3), 7-3ln(3) ) ?

I'm just a bit confused graphically.

 

[xanthym]

so the point would be... ( ln(3), 7-3ln(3) ) ? <----- CORRECT

I'm just a bit confused graphically.

Your answer is CORRECT. If helpful, sketch the graph to better understand the math. (Or use a graphing calculator.)


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