Asst 18 Comp

course Mth 271

´³åh”ò©ÍÝ¿°ïÄwÎ[ø“Ï©xñ홳àïÛðÙassignment #018

018. `query 18

Applied Calculus I

05-05-2009

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15:58:24

** 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

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

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

g'(x) * f'((g(x))

-3x^2 * -6 / (x^3 -4)^3

18x^2 / (x^3 - 4)^3

self critique assessment: 2

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16:00:40

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 think I messed up my signs, but I multiplied 6 and 3, not added them. Should I have added them?

self critique assessment: 1

your f(x) function would have been 3 x^-1, not 3 x^-2. Otherwise OK.

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16:03:16

**** 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 -->

The equation of the tangent line is the derivative.

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

f(x) = 1 / sqrt x f'(x) = 1 / 2 *sqrt (x^3)

g'(x) * f'(g(x))

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

confidence assessment: 2

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

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

16:04:15

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

your (2x - 3) * 1 / 2 *sqrt (x^2 - 3x + 4)^3 is right, except the 1/2 should be -1/2.

You need to then evaluate the derivative at the given point, and find the equation of a straight line with that slope through the point, as is done below:

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 -->

I think I got lost in this equation. I don't quite understand the answer to this question. I can't seem to find where I messed up.

self critique assessment: 1

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

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

16:08:40

**** 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 -->

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

f(n) = .25 * sqrt(n)

g'(n) = n + 5

f'(n) = 1 / 2 * sqrt (n)

Be careful with signs of grouping.

Should be 1 / (2 * sqrt (n)), and should be multiplied by .25 to give you

f ' (n) = .25 * 1 / (2 * sqrt (n)), or 1 / (8 sqrt(n)).

g'(x) * f'((g(x))

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

(n + 5) / 2 * swrt( .5n^2 + 5n + 25)

At P' = (12000)

12000 + 5 / 2* sqrt ( .5 (12000^2) + 5(12000) + 25

12005 / 2 * sqrt ( 72000000 + 60000 + 25

12005 / 2 * sqrt( 72060025

12005 / 2 * 8488.82

12005 / 16977.64

.71

confidence assessment: 1

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

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

16:09:28

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 -->

Again I messed up the derivative but I don't know where. I feel comfortable with this process, but I keep messing up, but I can't pinpont where.

self critique assessment: 1

see my note inserted into your solution

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

`ht52

Asst 18 Comp

course Mth 271

´³åh”ò©ÍÝ¿°ïÄwÎ[ø“Ï©xñ홳àïÛðÙassignment #018

018. `query 18

Applied Calculus I

05-05-2009

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

15:58:24

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

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

RESPONSE -->

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

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

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

g'(x) * f'((g(x))

-3x^2 * -6 / (x^3 -4)^3

18x^2 / (x^3 - 4)^3

self critique assessment: 2

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

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

16:00:40

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 -->

I think I messed up my signs, but I multiplied 6 and 3, not added them. Should I have added them?

self critique assessment: 1

your f(x) function would have been 3 x^-1, not 3 x^-2. Otherwise OK.

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

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

16:03:16

**** 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 -->

The equation of the tangent line is the derivative.

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

f(x) = 1 / sqrt x f'(x) = 1 / 2 *sqrt (x^3)

g'(x) * f'(g(x))

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

confidence assessment: 2

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

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

16:04:15

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

your (2x - 3) * 1 / 2 *sqrt (x^2 - 3x + 4)^3 is right, except the 1/2 should be -1/2.

You need to then evaluate the derivative at the given point, and find the equation of a straight line with that slope through the point, as is done below:

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 -->

I think I got lost in this equation. I don't quite understand the answer to this question. I can't seem to find where I messed up.

self critique assessment: 1

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

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

16:08:40

**** 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 -->

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

f(n) = .25 * sqrt(n)

g'(n) = n + 5

f'(n) = 1 / 2 * sqrt (n)

Be careful with signs of grouping.

Should be 1 / (2 * sqrt (n)), and should be multiplied by .25 to give you

f ' (n) = .25 * 1 / (2 * sqrt (n)), or 1 / (8 sqrt(n)).

g'(x) * f'((g(x))

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

(n + 5) / 2 * swrt( .5n^2 + 5n + 25)

At P' = (12000)

12000 + 5 / 2* sqrt ( .5 (12000^2) + 5(12000) + 25

12005 / 2 * sqrt ( 72000000 + 60000 + 25

12005 / 2 * sqrt( 72060025

12005 / 2 * 8488.82

12005 / 16977.64

.71

confidence assessment: 1

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

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

16:09:28

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 -->

Again I messed up the derivative but I don't know where. I feel comfortable with this process, but I keep messing up, but I can't pinpont where.

self critique assessment: 1

see my note inserted into your solution

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

`ht52

Asst 18 Comp

course Mth 271

´³åh”ò©ÍÝ¿°ïÄwÎ[ø“Ï©xñ홳àïÛðÙassignment #018

018. `query 18

Applied Calculus I

05-05-2009

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

15:58:24

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

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

RESPONSE -->

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

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

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

g'(x) * f'((g(x))

-3x^2 * -6 / (x^3 -4)^3

18x^2 / (x^3 - 4)^3

self critique assessment: 2

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

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

16:00:40

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 -->

I think I messed up my signs, but I multiplied 6 and 3, not added them. Should I have added them?

self critique assessment: 1

your f(x) function would have been 3 x^-1, not 3 x^-2. Otherwise OK.

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

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

16:03:16

**** 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 -->

The equation of the tangent line is the derivative.

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

f(x) = 1 / sqrt x f'(x) = 1 / 2 *sqrt (x^3)

g'(x) * f'(g(x))

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

confidence assessment: 2

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

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

16:04:15

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

your (2x - 3) * 1 / 2 *sqrt (x^2 - 3x + 4)^3 is right, except the 1/2 should be -1/2.

You need to then evaluate the derivative at the given point, and find the equation of a straight line with that slope through the point, as is done below:

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 -->

I think I got lost in this equation. I don't quite understand the answer to this question. I can't seem to find where I messed up.

self critique assessment: 1

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

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

16:08:40

**** 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 -->

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

f(n) = .25 * sqrt(n)

g'(n) = n + 5

f'(n) = 1 / 2 * sqrt (n)

Be careful with signs of grouping.

Should be 1 / (2 * sqrt (n)), and should be multiplied by .25 to give you

f ' (n) = .25 * 1 / (2 * sqrt (n)), or 1 / (8 sqrt(n)).

g'(x) * f'((g(x))

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

(n + 5) / 2 * swrt( .5n^2 + 5n + 25)

At P' = (12000)

12000 + 5 / 2* sqrt ( .5 (12000^2) + 5(12000) + 25

12005 / 2 * sqrt ( 72000000 + 60000 + 25

12005 / 2 * sqrt( 72060025

12005 / 2 * 8488.82

12005 / 16977.64

.71

confidence assessment: 1

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

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

16:09:28

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 -->

Again I messed up the derivative but I don't know where. I feel comfortable with this process, but I keep messing up, but I can't pinpont where.

self critique assessment: 1

see my note inserted into your solution

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

`ht52