open qa 20

#$&*

course Mth 271

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

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Your solution:

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

2x - 3y^2y’ = 0

-3y^2y’ = -2x

3y^2y’ = 2x

Y’ = 2x/3y^2

Y= = 2 * 2/3 * 1^2 = 4/3

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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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:

d/dx (x^2 - y^3) = 2x - 3y^2y’

y’ = 2x/3y^2

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):

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Self-critique Rating:

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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

p^2 * 2x = 500 - x

p^2 2x +x - 500 = 0

2p2x + 2x’p^2 + x’ -0 = 0

4 px + 2x’ p^2 + x’ = 0

X’ (2p^2 + 1) = -4px

X’ = - 4px/(2p^2 +1)

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):

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Self-critique rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

p^2 * 2x = 500 - x

p^2 2x +x - 500 = 0

2p2x + 2x’p^2 + x’ -0 = 0

4 px + 2x’ p^2 + x’ = 0

X’ (2p^2 + 1) = -4px

X’ = - 4px/(2p^2 +1)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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) **

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

p^2 * 2x = 500 - x

p^2 2x +x - 500 = 0

2p2x + 2x’p^2 + x’ -0 = 0

4 px + 2x’ p^2 + x’ = 0

X’ (2p^2 + 1) = -4px

X’ = - 4px/(2p^2 +1)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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) **

"

Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

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