Asst 18 query

course Mth 271

²¹Ñ„ÄyÚÁü—¶ÌwIÿ¸Ëƒ´ûËßÁü©¹assignment #018

018. `query 18

Applied Calculus I

11-15-2008

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16:54:39

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

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RESPONSE -->

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

first I rewrite the function as

f(x) = 3(x^3-4)^-2

then I use the general power rule

f'(x) = -6(x^3 - 4)^-3 (3x^2)

f'(x) = -18x^2 / (x^3 - 4)^3

self critique assessment: 3

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16:57:43

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 -->

I believe I got the question right because in the answer given here, the derivative of the f function does not multiply the constant by the power of z and the power of z stays -2 instead of becoming -3.

self critique assessment: 2

The given solutoin was for f(x) = 3/(x^3-4).

If f(x) is in fact f(x) = 3/(x^3-4)^2 then your solution is correct.

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18:16:46

**** 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 -->

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

I rewrite it as

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

Now I use the general power rule

f'(x) = -.5(x^2-3x+4)^(-3/2) (2x-3)

f'(x) = (-x+1.5) / (x^2-3x+4)

Now I substitute 3 for x to find the slope.

f'(3) = (-3 + 1.5) / (3^2 - 3(3) + 4) = -.19 approx.

Finally, I use the point slope form to find the equation of the tangent line.

y - 1/2 = -.19(x - 3)

y = -.19x + 1.07

confidence assessment: 2

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18:18:55

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 -->

I worked everything right. I probably should have left my answer as a fraction. That way I would have gotten an exact answer instead of a rounded answer.

self critique assessment: 2

The fraction is more accurate, the approximation easier to compare to other functions. Both have their advantages, but when in doubt, it's usually best to use the exact answer. If you approximate you lose information.

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18:35:31

**** 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 = .25 `sqrt(.5n^2+5n+25)

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

P' = (.125n + .625) / `sqrt(.5n^2+5n+25)

When the population is 12,000

P'(12) = [.125(12) + .625] / `sqrt[.5(12)^2 + 5(12) + 25]

P'(12) = (1.5 + .625) / `sqrt(72 + 60 + 25)

P'(12) = 2.125 / `sqrt(157) = .170 approx.

So at a population of 12,000, the pollution is increasing at a rate of .170 parts per million / thousand of population.

confidence assessment: 2

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