Asst 20 query

course Mth 271

????}?????assignment #020

020. `query 20

Applied Calculus I

11-24-2008

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22:17:29

2.7.16 dy/dx at (2,1) if x^2-y^3=3

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RESPONSE -->

x^2 - y^3 = 3

I take the derivative using implicit differentiation.

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

Now I solve for dy/dx

3y^2 dy/dx = 2x

dy/dx = 2x / 3y^2

Next, I evaluate for the point (2, 1)

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

confidence assessment: 3

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22:20:46

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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RESPONSE -->

self critique assessment: 3

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22:25:27

2.7.30 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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RESPONSE -->

x^2 - y^3 = 0

I take the derivative using implicit differentiation.

2x - 3y^2 y' = 0

3y^2 y' = 2x

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

To find the slope of the line, I evaluate the derivative at the given point (-1, 1)

y' = 2 * (-1) / (3 * 1^2)

y' = -2/3

confidence assessment: 3

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22:25:49

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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RESPONSE -->

self critique assessment: 3

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22:45:49

2.7.36 p=`sqrt( (500-x)/(2x))

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RESPONSE -->

p=`sqrt( (500-x)/(2x))

I begin by squaring both sides

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

Now I take the derivative, using implicit differentiation and the quotient rule

(p^2)' = [(500-x)' (2x) - (500-x) (2x)'] / (2x)^2

2p = [-2x dx/dp - 2 dp/dx (500-x)] / 4x^2

Should be

2p = [-2x dx/dp - 2 dx/dp (500-x)] / 4x^2

I multiply both sides by 4x^2

-2x dx/dp - 2 dp/dx (500-x) = 8px^2

-2x dx/dp - 1000 dp/dx + 2x dx/dp = 8px^2

The 2x dx/dp and -2x dx/dp add up to zero leaving me with

-1000 dp/dx = 8px^2

dp/dx = -8px^2 / 1000 = -px^2 / 125

confidence assessment: 2

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23:00:01

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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RESPONSE -->

I see that it probably would have been easier to set everything equal to zero at the beginning. Still, I thought I worked the problem right. I have looked over my work several times and I can't see what went wrong that caused me to get a different answer.

self critique assessment: 2

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Very good. See my note below related to that last question.

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

2 p = ( (500 - x) / (2x) ) ' * dx/dp

2 p = (- 2 x - 2 * (500 - x) ) / (4 x^2) * dx/dp

2 p = -1000 / (4 x^2) dx/dp

dx/dp = -p x^2 / 125

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

2x p^2= 500 - x

2x p^2 + x - 500 = 0

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

...

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

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

2x p^2= 500 - x

2x p^2 + x = 500

x * (2 p^2 + 1) = 500

x = 500 / (2 p^2 + 1) so

dx/dp = -500 ( 4 p + 1) / (2 p^2 + 1)^2

Since x = 500 / ( 2 p^2 + 1) this matches my solution

dx/dp = -4 x p / (2 p^2 + 1), i.e.,

dx/dp = -4 ( 500 / (2 p^2 + 1) ) * p / (2 p^2 + 1) = -2000 p / (p^2 + 1)^2.

That's three different solutions, two of which nearly agree, and both of which are not in agreement with yours, so I suspect you have an error. However I can't spot it without writing it down on paper.

I'll get back to you as soon as I have a chance to do this, but I need to post this now.