Assignment 20

course Mth 271

020. `query 20

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Question: `q2.7.16 (was 2.7.10) dy/dx at (2,1) if x^2-y^3=3

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Your solution:

derivative of x^2 with respect to x is 2 x.

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.

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

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

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

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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:ok

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Question: `q2.7.28 (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:

The derivative is

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

Solve

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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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):

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Self-critique Rating:

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Question: `q2.7.42 (was 2.7.36) p=`sqrt( (500-x)/(2x))

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Your solution:

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

2x p^2 = 500-x and

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

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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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):

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Self-critique Rating:ok

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