# Easy calculus... :sWatch

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#1
Really don't know why I can't do this , 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?
0
14 years ago
#2
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)
0
14 years ago
#3
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
0
#4
(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...
0
#5
Ok, so I've integrated to find area under curve. How do i find area between tangent and curve though?

Thanks...
0
14 years ago
#6
(Original post by dinkymints)
Ok, so I've integrated to find area under curve. How do i find area between tangent and curve though?

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.
0
#7
(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.
I thought it was that but then thought that was too simple... oh well, I'm just doing them like that anyway
0
14 years ago
#8
Did you get the question from a past paper? I could try and find the answer for you if you get it wrong.
0
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