QUERY 20

course MTH 271

11/10,1430

assignment #020020. `query 20

Applied Calculus I

11-08-2009

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23:21:12

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

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RESPONSE -->

x^2 - y^3 = 3

d/dx[x^2] - d/dx[y^3] = d/dx[3]

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

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

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

At (2,1): (2(2)) / (3(1^2)) = (4/3)

confidence rating: 2

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23:21:20

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 -->

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Self-critique Rating:ent: 3

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23:24:22

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

d/dx[x^2] - d/dx[y^3] = d/dx[0]

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

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

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

At (-1,1): (2(1)) / (3(1^2)) = (-2/3)

confidence rating: 2

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23:24:38

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 -->

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Self-critique Rating:ent: 3

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23:33:58

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

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RESPONSE -->

p= (500 - x)^(1/2) / (2x)^(1/2)

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

2xp^2=500 - x

2xp^2 + x - 500 = 0

d/dp [2xp^2] + d/dp[x] - d/dp[500] = d/dp[0]

2(x(2p) + dx/dp(p^2)) + dx/dp =0

4xp + 2p^2(dx/dp) + dx/dp =0

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

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

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

confidence rating: 2

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23:37:05

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 -->

My solution matches the given answer almost exactly. However, instead of having (2p^2 - 1) as a denominator, I found (2p^2 + 1). After going through the steps again, I still cannot find how there is a minus one. As the ( 500 - x) is moved to the opposite side of the equation, the x becomes positive and the 500 is subtracted. When taking the differential of x with regard to p, I found dx/dp as being added in the equation, not subtracted. Even after later factoring, I still did not have the dx/dp being subtracted or negative. I'm not sure exactly what I'm missing here.

Your solution is correct. The term is + x, and when we take the p derivative we get + dx/dp, not - dx/dp. That's an apparent typo on my part, which was then perpetuated through the rest of the given solution.

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Self-critique Rating:ent: 1"

&#This looks good. See my notes. Let me know if you have any questions. &#