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ilovealgebra

Member

Registered: 2006-10-02

Posts: 40

Implicit Differentiation

OK having problems with this one: Find the gradient of the normal to the curve 2xy ² -x²y³ =1 at the point (1,1)

Thanks in advance!

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Implicit Differentiation

Hi ilovealgebra;

I was doing a similar problem a couple of days ago. I found this then:
http://qaboard.cramster.com/calculus-topic-5-164162-cpi0.aspx
this should give you everything you need.

Last edited by bobbym (2009-05-02 18:54:27)

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In mathematics, you don't understand things. You just get used to them.
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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Implicit Differentiation

Or to explain how it works:

The normal to a curve always has a gradient of -1/m from the tangent, it is calculating the tangent normal that is the complex part.
Method 1 is the standard approach that you would be taught in school, method 2 is a much simpler way of doing it that relies on a different kind of differentiation that usually isn't taught at high-school level, atleast not at this stage.

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Method 1:

Remembering the chain rule:

in this case, we need to be able to differentiate a function of y, in terms of x

so wherever there is a function of y, differentiate with respect to y, and multiply by dy/dx
when you have as above, a function of x multiplied with a function of y, you must then use the product rule:

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Method 2:

without an explanation of why this is, it is immediately apparent that it is true with the worked example above, if you are not familiar with partial derivatives, they are found by differentiating with respect to one variable, whilst assuming that all other variables are true, (as oposed to implicit where we allow that all variables can vary)

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either way, once you have the derivitive, plug in your values of x and y, and find the negative reciprocal -1/m for the gradient of the normal

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Another interesting result of partial differentiation, is that the vector formed with the partial derivitives as the components gives you a direct equation for the normal to a curve (as a vector, not a line), in which case the normal to the curve in my worked example would be (non normalised)

in which case it's gradient would be dy/dx as

which you can immediately see is -1/m from dy/dx for the tangent

you could also turn this around, and use it as an argument for why the gradient of the curve is given by:

Last edited by luca-deltodesco (2009-05-02 21:41:28)

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