query 18

course mth 271

駯fw\c¬assignment #018

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

018. `query 18

Applied Calculus I

11-19-2007

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

14:42:02

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

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

RESPONSE -->

f(x)= 3/z g(x)=x^3-4

1st find f'(x) using the quotient rule

f'(x)= (z(0)-3(1))/z^2

f'(x)= -3/z^2

using the chain rule:

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

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

self critique assessment: 3

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

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

14:42:15

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 -->

yup

self critique assessment: 3

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

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

14:47:58

**** 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 -->

f(x)= 1/sqroot(z)

g(x)=x^2-3x+4

1) find f'(x) using quotient rule

sqroot((z)(0)-(1)(1/2z^-1/2))/ ((sqrootz)^2)

f'(x)=(-1/2z^-1/2)/(z)

f'(x)= -1/2z^-3/2

2)find f'(g(x)) using the chain rule

g'(x)= 2x-3

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

3) plug in x=3

f'(3)= -3/16

4) use pt slope form

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

y= -3/16x +17/16

confidence assessment: 3

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

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

14:48:07

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 -->

yup

self critique assessment: 3

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

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

14:49:12

**** 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 -->

im not sure what the two f(x) are...

f(x)=.25sqroot(z)

g(x)=.5n^2+5n+25

then use the product rule to get f'(x)?? doesnt sound right

confidence assessment: 0

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

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

14:49:49

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 -->

okay im not seeing where the derivative is coming from

what is f(x) and what is g(X) and which rule was used?

self critique assessment: 0

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

You've got it, but the expression is a function of n. You should have

f(z) = sqrt(z)

g(n) = .5 n^2 + 5 n + 25.

So f ' (z) = 1 / (2 sqrt(z)), g ' (n) = n + 5 and

(f(g(n))) ' = g ' (n) * f ' (g(n)) = (n + 5) * 1 / (2 sqrt(g(n)) = (n + 5) / (2 sqrt( .5 n^2 + 5 n + 25)).

*************************

&#

Good responses. See my notes and let me know if you have questions. &#