Assignment 20

course MTH271

10/31/09 4:23pm

Question: `q 2b 7th edition 2.7.16 (was 2.7.10) dy/dx at (2,1) if x^2-y^3=3YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating:

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Given Solution:

`a The derivative of x^2 with respect to x is 2 x.

The derivative of y^3 with respect to x is 3 y^2 dy/dx. You can see this by realizing that since y is implicitly a function of x, y^3 is a composite function: inner function is y(x), outer function f(z) = z^3. So the derivative is y'(x) * 3 * f(y(x)) = dy/dx * 3 * y^3.

So the derivative of the equation is

2 x - 3 y^3 dy/dx = 0, giving

3 y^2 dy/dx = 2 x so

dy/dx = 2 x / ( 3 y^2).

At (2,1), we have x = 2 and y = 1 so

dy/dx = 2 * 2 / (3 * 1^2) = 4/3. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q 3b 7th edition 2.7.28 (was 2.7.22) slope of x^2-y^3=0 at (-1,1)

What is the desired slope and how did you get it?

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Your solution: 2x-3y^3(y')=0

-3y^3(y')= -2x

y'= (2x)/(3y^3)

y' = 2/3

confidence rating:

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Given Solution:

`a The derivative of the equation is

2 x - 3 y^2 dy/dx = 0. Solving for dy/dx we get

dy/dx = 2x / (3 y^2).

At (-1,1) we have x = -1 and y = 1 so at this point

dy/dx = 2 * -1 / (3 * 1^2) = -2/3. **

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Self-critique (if necessary):Well you give 2 different points for x -1 and 1

The statement above did use (1, 1), and your solution was correct for that point.

Should be clear why y ' = -2/3 is correct for (-1, 1).

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Self-critique Rating:3

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Question: `q 6 7th edition 2.7.42 (was 2.7.36) p=`sqrt( (500-x)/(2x))

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Your solution: p^2= (500-x)/(2x)

2xp^2=500-x

2xp^2 +x -500=0

confidence rating:

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Given Solution:

`a You could apply implicit differentiation to the present form, and that would work but it would be fairly messy.

You have lots of choices for valid ways to rewrite the equation but I would recommend squaring both sides and getting rid of denominators. You get

p^2 = (500-x) / (2x) so

2x p^2 = 500-x and

2x p^2 + x - 500 = 0.

You want dx/dp so take the derivative with respect to p:

2x * 2p + 2 dx/dp * p^2 - dx / dp = 0

(2 p^2 - 1) dx/dp = - 4 x p

dx / dp = -4 x p / (2p^2 - 1) **

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Self-critique (if necessary):could you explain that differiant I not sure what happens to the middle x

Remember how implicit differentiation works, with on variable being considered as a function of the other. In this case x is considered to be a function of p.

One way to see it:

if ' represents derivative with respect to p, then

(2 x p^2) ' = 2 ( x p^2) ' = 2 ( x ' p^2 + x * (p^2)' ) = 2 ( x ' p^2 + x * (2 p) ) = 2 x ' p^2 + 4 x p.

So the equation

2x p^2 + x - 500 = 0 gives us

(2 x p^2) ' + x ' + 0 = 0, or

2 x ' p^2 + 4 x p + x ' = 0.

If we use d/dp notation, this is

2 dx/dp + p^2 + dx/dp = 0.

Remember that the derivative is with respect to p, not x, so x ' is x ' = dx/dp, not x ' = 1 (as it would be if we were taking a derivative with respect to x).

Another way to notate it:

The equation is 2 x p^2 = 500 - x.

We take the derivative of both sides with respect to p. So the derivative of x is dx/dp. The derivative of p^2 with respect to p is just 2 p.

The derivative of x p^2 is, by the product rule,

derivative of x p^2 = (derivative of x) * p^2 + x * (derivative of p^2), which works out to

dx/dp * p^2 + x * (2 p).

Of course, that means that the derivative of 2 x p^2 is 2 ( dx/dp * p^2 + x * (2 p) ) = 2 dx/dp * p^2 + 4 x p.

So the derivative with respect to p of 2 x p^2 + x - 500 is

(2 dx/dp * p^2 + 4 x p) + dx/dp + 0 or just

(2 dx/dp * p^2 + 4 x p) + dx/dp,

and our equation is

2 dx/dp * p^2 + 4 x p + dx/dp = 0.

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Self-critique Rating:1

&#Good responses. See my notes and let me know if you have questions. &#