Assignment 3 

course Mth 272

THIS actually consists of Assignment of 2 and 3 because the SEND file is from the same day.

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.

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lwѭƽʢczUw

assignment #003

{h׌İ}˭ǚ

Applied Calculus I

06-04-2006

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21:13:52

4.5.10 (was 4.4.10) find the derivative of ln(1-x)^(1/3)

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RESPONSE -->

1/3 ln (1-x) ---> 1/3 ( 1/(1-x) ) so 1 / [3(1-x)]

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21:20:44

** The function is of the form ln(u), so the derivative is 1/u * u', or ln(u) * du/dx. The function u is (1-x)^(3/2).

The derivative of this function is u' = du/dx = -1 * 3/2 * (1-x)^(1/2) = -3/2 (1-x)^(1/2).

Thus the derivative of the original function is

1/u du/dx =

1 / [(1-x)^(3/2) ] * [-3/2 (1-x)^(1/2)] =

-3/2 (1-x)^(1/2) (1-x)^(-3/2) =

-3/2 (1-x)^-1 =

-3 / [ 2 (1-x) ]

ALTERNATIVE SOLUTION:

Note that ln(1-x)^(1/3) = 1/3 ln(1-x)

The derivative of ln(1-x) is u ' * 1/u with u = 1-x. It follows that u ' = -1 so the derivative of ln(1-x) is -1 * 1/(1-x) = -1/(1-x).

The derivative of 1/3 ln(1-x) is therefore 1/3 * -1/(1-x) = -1 / [ 3(1-x) ].**

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RESPONSE -->

ok

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions.

The detail you missed was the u ' = -1 part, which gave you the wrong sign in your answer. Be sure you note every detail of the given solution.

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21:24:31

4.5.25 (was 4.4.24) find the derivative of ln( (e^x + e^-x) / 2)

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RESPONSE -->

ln (e^x + e^-x) - ln 2 because when use of division occurs the use of subtraction is involved.

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21:25:25

** the derivative of ln(u) is 1/u du/dx; u = (e^x + e^-x)/2 so du/dx = (e^x - e^-x) / 2.

The term - e^(-x) came from applying the chain rule to e^-x.

The derivative of ln( (e^x + e^-x) / 2) is therefore

[(e^x - e^-2) / 2 ] / ] [ (e^x + e^-x) / 2 ] = (e^x - e^-x) / (e^x + e^-x).

This expression does not simplify, though it can be expressed in various forms (e.g., (1 - e^-(2x) ) / ( 1 + e^-(2x) ), obtained by dividing both numerator and denominator by e^x.).

ALTERNATIVE SOLUTION:

ln( (e^x + e^-x) / 2) = ln( (e^x + e^-x) ) - ln(2).

the derivative of e^(-x) is - e^(-x) and ln(2) is constant so its derivative is zero.

So you get

y ' = (e^x - e^-x)/(e^x + e^-x). **

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RESPONSE -->

[(e^x - e^-2) / 2 ] / ] [ (e^x + e^-x) / 2 ] = (e^x - e^-x) / (e^x + e^-x).

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21:27:08

4.5.30 (was 4.4.30) write log{base 3}(x) in exp form

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RESPONSE -->

ln x / ln 3 = 1/ln3 (d/dx ln x) = 1/ x ln 3

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21:27:33

** the exponential form of y = log{base 3}(x) is x = 3^y, which I think was the question -- you can check me on that and let me know if I'm wrong **

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RESPONSE -->

ok

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21:28:26

** Write 25^u where u = 2x^2. So du/dx = 4x.

The derivative of a^x is a^x * ln(a). So the derivative of 25^u with respect to x is

du / dx * ln(25) * 25^u = 4x ln(25) * 25^u = 4x ln(25) * 25^ (2 x^2).

Evaluating this for x = -1/2 you get

4 * (-1/2) ln(25) * 25^(2 * (-1/2)^2 ) = -2 ln(25) * 25^(1/2) = -2 ln(25) * 5 = -10 ln(25) = -20 ln(5) = -32.189 approx.

So the tangent line is a straight line thru (-1/2, 5) and having slope -20 ln(5). The equation of a straight line with slope m passing thru (x1, y1) is

(y - y1) = m ( x - x1) so the slope of the tangent line must be

y - 5 =-20 ln(5) ( x - (-1/2) ) or

y - 5 = -20 ln(5) x - 10 ln(5). Solving for y we get

y = -20 ln(5) x - 10 ln(5).

A decimal approximation is

y = -32.189x - 11.095

ALTERNATIVE SOLUTION:

A straight line has form y - y1 = m ( x - x1), where m is the slope of the graph at the point, which is the value of the derivative of the function at the point. So you have to find the derivative of 25^(2x^2) then evaluate it at x = -1/2.

The derivative of a^x is ln(a) * a^x. The derivative of 25^z would therefore be ln(25) * 25^z. The derivative of 25^(2 x^2) would be found by the chain rule with f(z) = 25^z and g(x) = 2 x^2. The result is g ' (x) * f ' (g(x)) = 4 x * ln(25) * 25^(2x^2). Evaluating at x = -1/2 we get -2 ln(25) * 25^(1/2) = -10 ln(25).

Now we use the ponit-slope form of the equation of a straight line to get (y - 5) = -10 ln(25) * (x - (-1/2) ) and simplify. **

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RESPONSE -->

ok

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21:31:09

4.5.59 (was 4.4.59) dB = 10 log(I/10^-16); find rate of change when I=10^-4

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RESPONSE -->

10 log 10^-4 / 10^-16 = 10

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21:31:33

** This function is a composite; the inner function is I / 10^-16, which has derivative 1/10^-16 = 10^16. So the derivative is

dB' = dB / dI = 10 ( 10^16 * / ln(10) ) / (I / 10^-16) = 10 / [ ln(10) * I ].

Alternatively, 10 log(I / 10^-16) = 10 (log I - log(10^-16) ) = 10 log I + 160; the derivative comes out the same with no need of the chain rule.

Plugging in I = 10^-4 we get rate = 10 / [ ln(10) * 10^-4 ] = 10^5 / ln(10), which comes out around 40,000 (use your calculator to get the accurate result. **

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RESPONSE -->

find the derivative not just solve the equation.

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21:33:05

4.5.60 (was 4.4.60) T = 87.97 + 34.96 ln p + 7.91 `sqrt(p); find rate of change

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RESPONSE -->

get the derivative of ln and then solve with appropriate p value.

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21:33:30

** The derivative with respect to p of ln p is 1 / p and the derivative with respect to p of sqrt(p) is 1 / (2 sqrt(p)). The derivative of the constant 89.97 is zero so

dT/dp = 34.96 * 1/p + 7.91 * 1 / (2 sqrt(p)) = 34.96 / p + 3.955 / sqrt(p). **

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RESPONSE -->

dT/dp = 34.96 * 1/p + 7.91 * 1 / (2 sqrt(p)) = 34.96 / p + 3.955 / sqrt(p).

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"

Be sure you see my notes on self-critique.

You appear to be a little rusty on the chain rule, and perhaps on other rules of differentiation. You need to be very good with differentiation when we start integration in Chapter 5.