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#1 2007-11-30 16:09:45
- sarah.nz
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- Registered: 2007-11-30
- Posts: 3
differentiation...
Find the equation of the normal to y=x^4 -4x^3 at the point for which x=1/2
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#2 2007-11-30 16:44:49
- Identity
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- Registered: 2007-04-18
- Posts: 934
Re: differentiation...
We have
At x = 1/2, we have
The y-coordinate is:
The gradient of the tangent is:
So the gradient of the normal is
(negative reciprocal).So the line we want passes through (1/2, -7/16) and has a gradient of 2/5.
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#3 2007-11-30 16:55:16
- sarah.nz
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- Registered: 2007-11-30
- Posts: 3
Re: differentiation...
Ahhh.. thank you, i did all of that except i used the gradient of the tangent for the equation of the normal!! stupid me =p
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