Assignment 20 Query

course Mth 271

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17:03:09

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

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RESPONSE -->

x^2-y^3=3

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

dy/dx = 2x/3y^2

At (2,1) dy/dx = 4/3

confidence assessment: 3

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17:03:13

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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17:03:24

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 -->

dy/dx = 2x/3y^2

at (-1,1) dy/dx = -2/3

confidence assessment: 3

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17:03:27

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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17:03:36

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

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RESPONSE -->

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)

confidence assessment: 3

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17:03:38

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 -->

self critique assessment: 3

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Very good work. Let me know if you have questions. &#