Assignment 18 Query

course Mth 271

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16:58:50

** Query problem 2.5.48 der of 3/(x^3-4) **** What is your result?

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RESPONSE -->

3/(x^3-4) = 3(x^3-4)^-1

3*-1(x^3-4)^-2 * 3x^2

-9x^2(x^3-4)^-2

f'(x) = -9x^2/(x^3-4)^2

self critique assessment: 3

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16:58:53

This function can be expressed as f(g(x)) for g(x) = x^3-4 and f(z) = 3 / z. The 'inner' function is x^3 - 4, the 'outer' function is 1 / z.

So f'(z) = -3 / z^2 and g'(x) = 3x^2.

Thus f'(g(x)) = -3/(x^3-4)^2 so the derivative of the whole function is

[3 / (x^3 - 4) ] ' = g'(x) * f'(g(x)) = 3x^2 * (-3/(x^3-4)^2) = -9 x^2 / (x^3 - 4)^2.

DER**

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RESPONSE -->

self critique assessment: 3

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16:59:07

**** Query problem 2.5.66 tan line to 1/`sqrt(x^2-3x+4) at (3,1/2) **** What is the equation of the tangent line?

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RESPONSE -->

1/`sqrt(x^2-3x+4) = 1(x^2-3x+4)^-1/2

s'(x) = -1/2(x^2-3x+4)^-3/2 * (2x-3)

s'(3) = -3/16

y-1/2 = -3/16(x-3)

y = -3/16x + 17/16

confidence assessment: 3

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16:59:09

The derivative is (2x - 3) * -1/2 * (x^2 - 3x + 4) ^(-3/2) .

At (3, 1/2) we get -1/2 (2*3-3)(3^2- 3*3 + 4)^(-3/2) = -1/2 * 3 (4)^-(3/2) = -3/16.

The equation is thus ( y - 1/2) = -3/16 * (x - 3), or y = -3/16 x + 9/16 + 1/2, or y = -3/16 x + 17/16.

DER**

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RESPONSE -->

self critique assessment: 3

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16:59:21

**** Query problem 2.5.72 rate of change of pollution P = .25 `sqrt(.5n^2+5n+25) when pop n in thousands is 12 **** At what rate is the pollution changing at the given population level?

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RESPONSE -->

p'(n) = .25(.5n^2+5n+25)^1/2

= .25*1/2(.5n^2+5n+25)^-1/2 (n+5)

= .125(.5n^2+5n+25)^-1/2 (n+5)

p'(12) = .16593

confidence assessment: 3

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16:59:23

The derivative is .25 [ (n + 5) * 1/2 * (.5 n^2 + 5 n + 25) ^(-1/2) )

= (n+5) / [ 8 `sqrt(.5n^2 + 5n + 25) ]

When n = 12 we get (12+5) / ( 8 `sqrt(.5*12^2 + 5 * 12 + 25) ) = 17 / 100 = .17, approx.

DER**

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RESPONSE -->

self critique assessment: 3

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