Help with CDs

course Mth 173

I got mny CDs. But I'm missing CD # 10. What do you want me to do.

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.

assignment #019

¡qíÈ•|·‰ÑéD“ñ\û‹Õá«³ÜðCÂô

Calculus I

07-24-2006

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11:36:06

Query problem 3.4.29 (3d edition 3.4.20) was 4.4.12 Derivative of `sqrt( (x^2*5^x)^3

What is the derivative of the given function?

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RESPONSE -->

this will also be (x^2 * 5^x)^(-1/3). so g(x) is x^2 * 5^x so g' (x) = 2x

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11:43:44

** The function is `sqrt( (x^2 * 5^x)^3 ) = (x^2 * 5^x)^(3/2).

This is of form f(g(x)) with g(x) = x^2 * 5^x and f(z) = z^(3/2). Thus when you substitute you get f(g(x)) = g(x)^(3/2) = (x^2 * 5^x)^(3/2).

(x^2 * 5^x) ' = (x^2)' * 5^x + x^2 * (5^x) ' =

2x * 5^x + x^2 ln 5 * 5^x =

(2x + x^2 ln 5) * 5^x.

`sqrt(z^3) = z^(3/2), so using w(x) = f(g(x)) with f(z) = z^(3/2) and g(x) = x^2 * 5^x we get

w ' = (2x + x^2 ln 5) * 5^x * [3/2 (x^2 * 5^x)^(1/2)] = 3/2 (2x + x^2 ln 5) * | x | * 5^(3/2 x).

Note that sqrt(x^2) is | x |, not just x, since the square root must be positive and x might not be. **

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RESPONSE -->

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11:48:09

Query problem 3.4.28 (3d edition 3.4.19) (was 4.4.20) derivative of 2^(5t-1).

What is the derivative of the given function?

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RESPONSE -->

g(x) = 5t + 1 so g'(x) = 5

f(x) = 2^x so f'(x) = ln(2) * 2^x

so g'(x) * f'(g(x)) = 5 * (ln(2) * 2^(5t + 1))

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11:48:37

This function is a composite. The inner function is g(x)=5t-1 and the outer function is f(z)=2^z.

f'(z)=ln(2) * 2^z.

g ' (x)=5

(f(g(t)) ' = g ' (t)f ' (g(t))=

5 ln(2) * 2^(5t-1).

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RESPONSE -->

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11:50:43

**** Query 3.4.68 (3d edition 3.4.56) y = k (x), y ' (1) = 2.

What is the derivative of k(2x) when x = 1/2?

What is the derivative of k(x+1) when x = 0?

{]What is the derivative of k(x/4) when x = 4?

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RESPONSE -->

this means that k = 2. so when x = 1/2 at k(2x). it would be 2 * 2(2 * 1/2) = 2*2 = 4.

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11:52:09

** We apply the Chain Rule:

( k(2x) ) ' = (2x) ' * k'(2x) = 2 k(2x).

When x = 1/2 we have 2x = 1.

k ' (1) = y ' (1) = 2 so

when x = 1/2

( k(2x) ) ' = 2 k(2 * 1/2) = 2 * k'(1) = 2 * 2 = 4.

(k(x+1)) ' = (x+1)' k ' (x+1) = k ' (x+1) so

when x = 0 we have

(k(x+1) ) ' = k ' (x+1) = k ' (1) = 2

(k(x/4)) ' = (x/4)' k'(x/4) = 1/4 * k'(x/4) so when x = 4 we have

(k(x/4))' = 1/4 * k'(x/4) = 1/4 k'(4/4) = 1/4 * 2 = 1/2. **

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RESPONSE -->

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11:56:19

Query 3.4.81 (3d edition 3.4.68). Q = Q0 e^(-t/(RC)). I = dQ/dt.

Show that Q(t) and I(t) both have the same time constant.

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RESPONSE -->

g(x) = -t / (RC) so g'(x) = -1 / (RC)

f(x) = e^x so f'(x) = e^x

So g'(x) * f'(g(x)) = (-QO / (RC)) * e^(-t / (RC))

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11:52:22

** We use the Chain Rule.

(e^(-t/(RC)))' = (-t/(RC))' * e^(-t/(RC)) = -1/(RC) * e^(-t/(RC)).

So dQ/dt = -Q0/(RC) * e^(-t/(RC)).

Both functions are equal to a constant factor multiplied by e^(-t/(RC)).

The time constant for both functions is therefore identical, and equal to RC. **

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RESPONSE -->

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11:52:28

Query problem 3.5.5 (unchanged since 3d edition) (formerly 4.5.6). What is the derivative of sin(3x)

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RESPONSE -->

g(x) is 3x so g'(x) = 3. f(x) is sin(x) So f'(x) = cos(x) So g'x * f'(g(x)) = 3 * cos(3x).

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11:52:30

** sin(3x) is the composite of g(x) = 3x, which is the 'inner' function (the first function that operates on the variable x) and the 'outer' function f(z) = sin(z).

Thus f(g(x)) = sin(g(x)) = sin(3x).

The derivative is (f (g(x) ) ' = g ' (x) * f' ( g(x) ).

g ' (x) = (3x) ' = 3 * x ' = 3 ', and

f ' (z) = (sin(z) ) ' = cos(z).

So the derivative is [ sin(3x) ] ' = ( f(g(x) ) ' = g ' (x) * f ' (g(x) ) = 3 * cos(3x). **

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RESPONSE -->

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11:52:33

Query problem 3.5.48 (3d edition 3.5.50) (formerly 4.5.36). Give the equations of the tangent lines to graph of y = sin(x) at x = 0 and at `pi/3

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RESPONSE -->

y - 0 = 1(x - 0) = y = x

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11:52:35

** At x = 0 we have y = 0 and y ' = cos(0) = 1.

The tangent line is therefore the line with slope 1 through (0,0), so the line is y - 0 = 1 ( x - 0) or just y = x.

At x = `pi/3 we have y = sin(`pi/3) = `sqrt(3) / 2 and y ' = cos(`pi/3) = .5.

Thus the tangent line has slope .5 and passes thru (`pi/3,`sqrt(3)/2), so its equation is

y - `sqrt(3)/2 = .5 (x - `pi/3)

y = .5 x - `pi/6 + `sqrt(3)/2. Approximating:

y - .87 = .5 x - .52. So

y = .5 x + .25, approx.

Our approximation to sin(`pi/6), based on the first tangent line:

The first tangent line is y = x. So the approximation at x = `pi / 6 is

y = `pi / 6 = 3.14 / 6 = .52, approximately.

Our approximation to sin(`pi/6), based on the second tangent line, is:

y = .5 * .52 + .34 = .60.

`pi/6 is equidistant from x=0 and x=`pi/3, so we might expect the accuracy to be the same whichever point we use.

The actual value of sin(`pi/6) is .5. The approximation based on the tangent line at x = 0 is .52, which is much closer to .5 than the .60 based on the tangent line at x = `pi/3.

The reason for this isn't too difficult to see. The slope is changing more quickly around x = `pi/3 than around x = 0. Thus the tangent line will more more rapidly away from the actual function near x = `pi/3 than near x = `pi/2. **

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RESPONSE -->

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11:52:38

Query 3.5.34 (3d edition 3.5.40). Der of sin(sin x + cos x)

What is the derivative of the given function and how did you find it?

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RESPONSE -->

g(x) is sin(x) + cos(x) so g'(x) = cos(x) - sin(x).

f(x) is sin(x) so f'(x) = cos(x).

so g'x * f'(g(x)) = (cos(x) - sin(x)) * cos(sin(x) + cos(x))

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11:52:40

The function y = sin( sin(x) + cos(x) ) is the composite of g(x) = sin(x) + cos(x) and f(z) = sin(z).

The derivative of the composite is g ' (x) * f ' (g(x) ).

g ' (x) = (sin(x) + cos(x) ) ' = cos(x) - sin(x).

f ' (z) = sin(z) ' = cos(z).

So g ' (x) * f ' (g(x)) = ( cos(x) - sin(x) ) * cos( sin(x) + cos(x) ).

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RESPONSE -->

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11:52:43

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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

RESPONSE -->

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

You did well overall. However be sure you completely self-critique when your solution doesn't coincide witht he given solution. Let me know if you have questions.

Help with CDs

course Mth 173

I got mny CDs. But I'm missing CD # 10. What do you want me to do.

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.

assignment #019

¡qíÈ•|·‰ÑéD“ñ\û‹Õá«³ÜðCÂô

Calculus I

07-24-2006

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

11:36:06

Query problem 3.4.29 (3d edition 3.4.20) was 4.4.12 Derivative of `sqrt( (x^2*5^x)^3

What is the derivative of the given function?

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

RESPONSE -->

this will also be (x^2 * 5^x)^(-1/3). so g(x) is x^2 * 5^x so g' (x) = 2x

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

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

11:43:44

** The function is `sqrt( (x^2 * 5^x)^3 ) = (x^2 * 5^x)^(3/2).

This is of form f(g(x)) with g(x) = x^2 * 5^x and f(z) = z^(3/2). Thus when you substitute you get f(g(x)) = g(x)^(3/2) = (x^2 * 5^x)^(3/2).

(x^2 * 5^x) ' = (x^2)' * 5^x + x^2 * (5^x) ' =

2x * 5^x + x^2 ln 5 * 5^x =

(2x + x^2 ln 5) * 5^x.

`sqrt(z^3) = z^(3/2), so using w(x) = f(g(x)) with f(z) = z^(3/2) and g(x) = x^2 * 5^x we get

w ' = (2x + x^2 ln 5) * 5^x * [3/2 (x^2 * 5^x)^(1/2)] = 3/2 (2x + x^2 ln 5) * | x | * 5^(3/2 x).

Note that sqrt(x^2) is | x |, not just x, since the square root must be positive and x might not be. **

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

RESPONSE -->

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

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

11:48:09

Query problem 3.4.28 (3d edition 3.4.19) (was 4.4.20) derivative of 2^(5t-1).

What is the derivative of the given function?

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

RESPONSE -->

g(x) = 5t + 1 so g'(x) = 5

f(x) = 2^x so f'(x) = ln(2) * 2^x

so g'(x) * f'(g(x)) = 5 * (ln(2) * 2^(5t + 1))

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

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

11:48:37

This function is a composite. The inner function is g(x)=5t-1 and the outer function is f(z)=2^z.

f'(z)=ln(2) * 2^z.

g ' (x)=5

(f(g(t)) ' = g ' (t)f ' (g(t))=

5 ln(2) * 2^(5t-1).

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

RESPONSE -->

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

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11:50:43

**** Query 3.4.68 (3d edition 3.4.56) y = k (x), y ' (1) = 2.

What is the derivative of k(2x) when x = 1/2?

What is the derivative of k(x+1) when x = 0?

{]What is the derivative of k(x/4) when x = 4?

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

RESPONSE -->

this means that k = 2. so when x = 1/2 at k(2x). it would be 2 * 2(2 * 1/2) = 2*2 = 4.

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

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

11:52:09

** We apply the Chain Rule:

( k(2x) ) ' = (2x) ' * k'(2x) = 2 k(2x).

When x = 1/2 we have 2x = 1.

k ' (1) = y ' (1) = 2 so

when x = 1/2

( k(2x) ) ' = 2 k(2 * 1/2) = 2 * k'(1) = 2 * 2 = 4.

(k(x+1)) ' = (x+1)' k ' (x+1) = k ' (x+1) so

when x = 0 we have

(k(x+1) ) ' = k ' (x+1) = k ' (1) = 2

(k(x/4)) ' = (x/4)' k'(x/4) = 1/4 * k'(x/4) so when x = 4 we have

(k(x/4))' = 1/4 * k'(x/4) = 1/4 k'(4/4) = 1/4 * 2 = 1/2. **

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RESPONSE -->

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11:56:19

Query 3.4.81 (3d edition 3.4.68). Q = Q0 e^(-t/(RC)). I = dQ/dt.

Show that Q(t) and I(t) both have the same time constant.

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

RESPONSE -->

g(x) = -t / (RC) so g'(x) = -1 / (RC)

f(x) = e^x so f'(x) = e^x

So g'(x) * f'(g(x)) = (-QO / (RC)) * e^(-t / (RC))

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

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11:52:22

** We use the Chain Rule.

(e^(-t/(RC)))' = (-t/(RC))' * e^(-t/(RC)) = -1/(RC) * e^(-t/(RC)).

So dQ/dt = -Q0/(RC) * e^(-t/(RC)).

Both functions are equal to a constant factor multiplied by e^(-t/(RC)).

The time constant for both functions is therefore identical, and equal to RC. **

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

RESPONSE -->

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11:52:28

Query problem 3.5.5 (unchanged since 3d edition) (formerly 4.5.6). What is the derivative of sin(3x)

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

RESPONSE -->

g(x) is 3x so g'(x) = 3. f(x) is sin(x) So f'(x) = cos(x) So g'x * f'(g(x)) = 3 * cos(3x).

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

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

11:52:30

** sin(3x) is the composite of g(x) = 3x, which is the 'inner' function (the first function that operates on the variable x) and the 'outer' function f(z) = sin(z).

Thus f(g(x)) = sin(g(x)) = sin(3x).

The derivative is (f (g(x) ) ' = g ' (x) * f' ( g(x) ).

g ' (x) = (3x) ' = 3 * x ' = 3 ', and

f ' (z) = (sin(z) ) ' = cos(z).

So the derivative is [ sin(3x) ] ' = ( f(g(x) ) ' = g ' (x) * f ' (g(x) ) = 3 * cos(3x). **

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RESPONSE -->

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11:52:33

Query problem 3.5.48 (3d edition 3.5.50) (formerly 4.5.36). Give the equations of the tangent lines to graph of y = sin(x) at x = 0 and at `pi/3

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RESPONSE -->

y - 0 = 1(x - 0) = y = x

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

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11:52:35

** At x = 0 we have y = 0 and y ' = cos(0) = 1.

The tangent line is therefore the line with slope 1 through (0,0), so the line is y - 0 = 1 ( x - 0) or just y = x.

At x = `pi/3 we have y = sin(`pi/3) = `sqrt(3) / 2 and y ' = cos(`pi/3) = .5.

Thus the tangent line has slope .5 and passes thru (`pi/3,`sqrt(3)/2), so its equation is

y - `sqrt(3)/2 = .5 (x - `pi/3)

y = .5 x - `pi/6 + `sqrt(3)/2. Approximating:

y - .87 = .5 x - .52. So

y = .5 x + .25, approx.

Our approximation to sin(`pi/6), based on the first tangent line:

The first tangent line is y = x. So the approximation at x = `pi / 6 is

y = `pi / 6 = 3.14 / 6 = .52, approximately.

Our approximation to sin(`pi/6), based on the second tangent line, is:

y = .5 * .52 + .34 = .60.

`pi/6 is equidistant from x=0 and x=`pi/3, so we might expect the accuracy to be the same whichever point we use.

The actual value of sin(`pi/6) is .5. The approximation based on the tangent line at x = 0 is .52, which is much closer to .5 than the .60 based on the tangent line at x = `pi/3.

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RESPONSE -->

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11:52:38

Query 3.5.34 (3d edition 3.5.40). Der of sin(sin x + cos x)

What is the derivative of the given function and how did you find it?

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

RESPONSE -->

g(x) is sin(x) + cos(x) so g'(x) = cos(x) - sin(x).

f(x) is sin(x) so f'(x) = cos(x).

so g'x * f'(g(x)) = (cos(x) - sin(x)) * cos(sin(x) + cos(x))

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

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11:52:40

The function y = sin( sin(x) + cos(x) ) is the composite of g(x) = sin(x) + cos(x) and f(z) = sin(z).

The derivative of the composite is g ' (x) * f ' (g(x) ).

g ' (x) = (sin(x) + cos(x) ) ' = cos(x) - sin(x).

f ' (z) = sin(z) ' = cos(z).

So g ' (x) * f ' (g(x)) = ( cos(x) - sin(x) ) * cos( sin(x) + cos(x) ).

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RESPONSE -->

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

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11:52:43

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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

RESPONSE -->

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

You did well overall. However be sure you completely self-critique when your solution doesn't coincide witht he given solution. Let me know if you have questions.