Easy calculus... :s Watch

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dinkymints
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#1
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Really don't know why I can't do this :confused:, any help would be much appreciated!!

Find the tangent to the curve y = x^3 - 3x^2 + 2x + 1 when x=2. Find where the tangent meets the curve again. Find the area between the curve and the tangent in this region.

I seem to have forgotten how to do even the first bit... ive found the eqn of the tangent (which I got as y = 2x-3) but setting that as equal to the eqn of the curve and rearranging to equal zero is giving me x^3 - 3x^2 -2 = 0 which if I'm not much mistaken doesn't give whole values of x.... is that right?
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nas7232
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part 1

y = x^3 - 3x^2 + 2x + 1

ok dy/dx f(x)

then substitute x into the equation [the 1 you just worked out [dy/dx]]
you now have the gradient when x=2

substitute x into f(x) to get y.

then subsitute x,y,m into
y-y1=m(x-x1)
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[email protected]
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firstly differentiate with respect to x
dy/dx = 3x^2 - 6x + 2
substitute x in to find the gradient
dy/dx at x=2 = 3(4) - 6(2) + 2 = 2
when x=2, y=8-3(4)+2(2)+1 = 1
substitute all this info into the equation
m(x-x1)=y-y1
2(x-2)=y-1
y= 2x -3
now equate the tangent line and the curve equations
2x - 3 = x^3 - 3x^2 + 2x + 1
x^3 - 3x^2 + 4 = 0 we know that 2 is solution, so (x-2) is a factor
(x-2)(x^2 - x - 2) = 0
(x-2)(x-2)(x+1)=0 => x=-1
argh carry on from here coz I dont wanna do the last part
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dinkymints
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(Original post by [email protected])
firstly differentiate with respect to x
dy/dx = 3x^2 - 6x + 2
substitute x in to find the gradient
dy/dx at x=2 = 3(4) - 6(2) + 2 = 2
when x=2, y=8-3(4)+2(2)+1 = 1
substitute all this info into the equation
m(x-x1)=y-y1
2(x-2)=y-1
y= 2x -3
now equate the tangent line and the curve equations
2x - 3 = x^3 - 3x^2 + 2x + 1
x^3 - 3x^2 + 4 = 0 we know that 2 is solution, so (x-2) is a factor
(x-2)(x^2 - x - 2) = 0
(x-2)(x-2)(x+1)=0 => x=-1
argh carry on from here coz I dont wanna do the last part
Thanks, I was just being really stupid thinking that (x^3 - 3x^2 + 2x + 1) - (2x-3) was x^3 - 3x^2 -2... :rolleyes:
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dinkymints
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Ok, so I've integrated to find area under curve. How do i find area between tangent and curve though? :confused:

Thanks...
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Chris87
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(Original post by dinkymints)
Ok, so I've integrated to find area under curve. How do i find area between tangent and curve though? :confused:

Thanks...
Find the area under the curve between the points of intersection and then find the area under the tangent and subtract the smallest from the largest. I think that's right. :confused:
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dinkymints
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(Original post by Chris87)
Find the area under the curve between the points of intersection and then find the area under the tangent and subtract the smallest from the largest. I think that's right. :confused:
I thought it was that but then thought that was too simple... oh well, I'm just doing them like that anyway
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Chris87
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Did you get the question from a past paper? I could try and find the answer for you if you get it wrong.
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