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C3-edexcel:coordinate geometry/logs question

Although there is no coordinate geometry in the c3 book this was the title i though appropiate for this qu, id appreciate any help:

f(x)=e^(2x-1), x greater than or equal to 0
The curve C with equation y=f(x) meets the y axis at P. The tangent to C at P crosses the x axis at Q.
Find too 3 dp the area of triangle POQ where O is the origin.
Id appreciate any help
P = (0, e^-1) by substituting x=0 into f(x)

Now to find the tangent at P we need to differentiate to find the gradient at x = 0 which I make it to be 2e^-1 since f'(x) = 2e^(2x-1)

The tangent is of the form y=mx+c

=> y = (2e^-1)x + e^-1 since c is where the tangent crosses the y-axis which we know already to be e^-1

Then to find Q,

(2e^-1)x + e^-1 = 0 since y=0 when the tangent crosses the x-axis

Solve to find x then you have all of the coordinates and should to be able to work out the area from there
Reply 2
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~NC~
OP
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ok thanks :smile: Ill try it from there :biggrin:
Reply 3
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~NC~
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why here => y = (2e^-1)x + e^-1 since c is where the tangent crosses the y-axis which we know already to be e^-1
 have you subbed in the gradient to be 2e-1 as opposed to 2e(2x-1)?
C is just the point where the tangent crosses the y-axis, and we want it to meet the curve f(x) when x=0 so we need c = e^-1 otherwise it wouldnt be a tangent to the curve at the point P
Reply 5
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~NC~
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I know why we need c as e-1, but why have you put m as 2e-1 as opposed to 2e(2x-1) im confused :frown:
~NC~
why here => y = (2e^-1)x + e^-1 since c is where the tangent crosses the y-axis which we know already to be e^-1
 have you subbed in the gradient to be 2e-1 as opposed to 2e(2x-1)?


The gradient of the curve at P is dy/dx at x = 0.

dy/dx = 2e^(2x-1)

So subbing x = 0, the gradient is 2e^-1
Reply 7
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~NC~
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Oooo i get you :biggrin:
Sorry :p: And thanks :biggrin:
Reply 8
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~NC~
OP
11
thankyou :biggrin: I got the answer :biggrin: :biggrin:
No problem :smile:

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