assn 2

course mth 174

01-31-2008......!!!!!!!!...................................

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00:06:45

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

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RESPONSE -->

Making this pertain to physics makes it alot easier to understand.

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00:21:14

Query Section 6.3 #8, ds / dt = -32 t + 100, s = 50 when t = 0

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RESPONSE -->

s' = 100 - 32t

s= 100t - 16t^2 + C

50 = 100(0) - 16(0)^2 + C

50 =C

s = 100(t) - 16t^2 + 50

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00:21:37

What is the solution satisfying the given initial condition?

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RESPONSE -->

s = 100t - 16t^2 +50

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00:24:09

What is the general solution to the differential equation?

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RESPONSE -->

The differential is the rate of the intial equation.

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00:30:07

Query Section 6.3 #14 water balloon from 30 ft, v(t) = -32t+40

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RESPONSE -->

I'll go forward to the questions about the question.

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00:57:21

How fast is the water balloon moving when it strikes the ground?

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RESPONSE -->

v(t) = 40 - 32t

y(t) = 40t - 16t^2 +C

30= 40(0) - 16(0)^2 + C

30=C

The balloon hits the ground shortly (very shortly) after 3 secs. so the velocity would be around 59.2 ft/s

** the antiderivative function is s(t) = -16 t^2 + 40 t + c. Since the building is 30 ft high you know that s(0) = 30.

Following the same method used in the preceding problem we get s(t) = - 16 t^2 + 40 t + 30.

The water balloon strikes the ground when s(t) = 0. This occurs when

-16 t^2 + 40 t + 30 = 0. Dividing by 2 we have

-8 t^2 + 20 t + 15 = 0. The quadratic formula gives us

t = [ -20 +- `sqrt( 400 + 480) ] / ( 2 * -8) or

t = 1.25 +- sqrt(880) / 16 or

t = 1.25 +- 29.7 / 16, approx. or

t = 1.25 +- 1.87 or

t = 3.12 or -.62.

The solution t = 3.12 gives us v(t) = v(3.12) = -32 * 3.12 + 40 = -60, approx.

So when it strikes the ground the balloon is moving downward at 60 feet / second.**

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00:59:55

How fast is the water balloon moving when it strikes the 6 ft person's head?

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RESPONSE -->

the balloon strikes the persons head at exactly 3sec. so the velocity would be 56ft/s.

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01:04:58

What is the average velocity of the balloon between the two given clock times?

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RESPONSE -->

the velocity at 1.5 seconds is 8 ft/s and the velocity at 3 seconds is 56ft/s so the average velocity would be 32 ft/s.

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01:11:46

What function describes the velocity of the balloon as a function of time?

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RESPONSE -->

the first given equation v(t) = 40 - 32t

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01:12:16

Query Section 6.4 #19 (#18 3d edition) derivative of (int(ln(t)), t, x, 1)

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RESPONSE -->

d / dx= ln(t) dt

y = 1/t

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01:14:52

What is the desired derivative?

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RESPONSE -->

y = 1/t

** In the following we'll use the format [int(f(t), t, c, x] to stand for 'the integral of f(t) with respect to t with lower limit c and upper limit x'.

The 2d Fundamental Theorem says that d/dx [ int(f(t)), t, c, x ] = f(x). When applying this Theorem you don't integrate anything.

The integral we are given has limits x (lower) and 1 (upper), and is therefore equal to -int(ln(t), t, 1, x). This expression is in the form of the Fundamental Theorem, with c = 1, and its derivative with respect to x is therefore - ln(x).

Note that this Theorem is simply saying that the derivative of an antiderivative is equal to the original function, just like the derivative of an antiderivative of a rate-of-depth-change function is the same rate function we start with. **

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01:26:54

The Second Fundamental Theorem applies to an integral whose upper limit is the variable with respect to which we take the derivative. How did you deal with the fact that the variable is the lower limit?

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RESPONSE -->

I remembered that the derivative of ln(x) is 1/x and I assumed that it would still be right because it was right before the new theorem.

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02:12:30

Why do we use something besides x for the integrand?

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RESPONSE -->

because it is a function of time and not of x and y since the variable is t?

** In the statement of the Theorem the derivative is taken with respect to the variable upper limit of the interval of integration, which is different from the variable of integration.

The upper limit and the variable of integration are two different variables, and hence require two different names. **

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02:34:18

Query Section 6.4 #26 (3d edition #25) derivative of (int(e^-(t^2),t, 0,x^3)

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RESPONSE -->

y' = e^t^2

Chain rule f'(g(x)) (g'(x))

y= 2 e^t^2 t

** If we were finding (int(e^-(t^2),t, 0,x) the answer would just be e^-(x^2) by the Second Fundamental Theorem.

However the upper limit on the integral is x^3. This makes the expression int(e^(-t^2),t,0,x^3) a composite of f(z) = (int(e^-(t^2),t, 0,z) and g(x) = x^3. Be sure you see that the composite f(g(x)) = f(x^3) would be the original expression int(e^(-t^2),t,0,x^3).

g'(x) = 3x^2, and by the Second Fundamental Theorem f'(z) = e^-(z^2). Thus the derivative is

g'(x) f'(g(x))= 3x^2 * e^-( (x^3)^2 ).

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02:34:36

What is the desired derivative?

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RESPONSE -->

2 e^t^2 t

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02:42:10

How did you apply the Chain Rule to this problem?

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RESPONSE -->

By breaking the equation into it's two parts e^t and t^2 and using the chain rule.

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02:48:02

Why was the Chain Rule necessary?

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RESPONSE -->

because the equation was complex it had two functions inside one another so they had to be ""chained""

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Your work looks good. See my notes. Let me know if you have any questions. &#