QUERY 18

course MTH 271

10/31,0400

assignment #018018. `query 18

Applied Calculus I

10-31-2009

......!!!!!!!!...................................

03:47:40

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

......!!!!!!!!...................................

RESPONSE -->

3/(x^3-4)

(0)(x^3 - 4) - (3)(3x^2) / (x^3 - 4)^2

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

------------------------------------------------

Self-critique Rating:ent: 3

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

......!!!!!!!!...................................

03:47:57

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**

......!!!!!!!!...................................

RESPONSE -->

------------------------------------------------

Self-critique Rating:ent: 3

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

......!!!!!!!!...................................

03:55:49

**** 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?

......!!!!!!!!...................................

RESPONSE -->

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

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

@(3, 1/2)

(-1/2)(2(3) -3)(3^2 - 3(3) + 4)^(-3/2)

(-1/2)(2(3) -3) / (3^2 - 3(3) + 4)^(3/2)

(-3/2) / 8

-3/16

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

y - 1/2 = -3/16x + 9/16

y=-3/16x + 17/16

confidence rating: 2

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

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

......!!!!!!!!...................................

03:55:57

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**

......!!!!!!!!...................................

RESPONSE -->

------------------------------------------------

Self-critique Rating:ent: 3

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

......!!!!!!!!...................................

04:05:31

**** 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?

......!!!!!!!!...................................

RESPONSE -->

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

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

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

P'= .25 [ .5(n+5) / ( .5n^2 + 5n + 25)^(1/2) ]

P'=(n + 5) / 8 (.5n^2 + 5n + 25)^(1/2)

n=12

P'(12) = (12 + 5) / 8(.5(12^2) + 5(12) + 25)^(1/2)

P'(12)= 17/100.23

P'(12)= .17

confidence rating: 2

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

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

......!!!!!!!!...................................

04:05:41

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**

......!!!!!!!!...................................

RESPONSE -->

------------------------------------------------

Self-critique Rating:ent: 3

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

"

&#Your work looks very good. Let me know if you have any questions. &#