Mth 174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I have been working on problems on practice tests and I wanted to make sure I was doing these correctly.
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1. Solve the equation (t + 7) dy/dt = y / .09, y(3) = 80
I tried this in 2 different ways. First way:
(t + 7) dy/dt = y / .09
dy/dt = (y/.09)/(t + 7)
integral of dy/ (y/.09)= integral of dt/(t + 7)
ln(y/.09)= ln (t + 7)
dy / (y/.09) does not integrate to ln(y/.09).
dy / (y / .09) = .09 * dy / y, which integrates to .09 ln(y).
This changes your subsequent steps.
y/.09 = t + 7
y = (t + 7) .09
80 = (3+7).09
= -79.1
Second Way:
(t + 7) dy/dt = y / .09
Separate variables into:
dy -y/.09 = dt/(t + 7)
you can't get (t + 7) dy/dt = y / .09 into this form; the algebra just won't work that way
your previous approach was correct; you just made that one error (of course that was a significant error, but not difficult to fix).
integrate both sides to get:
-y^2/2 * 1/.09y = ln(t + 7) + c
t + 7 = A e^-y^2/2 * 1/.09y
3 + 7 = A e^ -(80^2/2) * 1/.09 * 80
10 = A e^0 = A
t = 7 + 10e^-y^2/2 * 1/.09* y
2. Use the Taylor polynomial of a simpler function to find the Taylor polynomial of degree 4 for the function f(x) = (8x^2.5 - 7)^.5, expanding about an appropriate value of the variable.
I first used the T. Polynomial of (1 + x)^p
F4) = 1 + .5x + (.5(.5 - 1)x^2)/2! + (.5(.5 - 1)(.5- 2)x^3)/3!+...etc
F4 = 1 + .5x + .25x^2/2! + .375x^3/3! - .9375x^4/4!
I then substituted (8x^2.5-7)= x to get
F4 = 1 + .5 (8x^2.5-7)+ .25^(8x^2.5-7)2/2! + .375^(8x^2.5-7)3/3! - .9375^(8x^2.5-7)4/4!
Would I leave it this way or simplify further?
That would be fine.
However to get the form 1 + x you would need to substitute 8 x^2.5 - 8, so that 8 x^2.5 - 7 would be if the form 1 + x.
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