Assignment 2

course Mth 174

If there are any tips to helping me with 6.4 i would like to know.

|UޘLWassignment #002

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Physics II

06-05-2008

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12:20:06

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

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RESPONSE -->

No real surprises. This assignment did not ask me any questions, other than that in the textbook. But I am still in need of practice with interpreting and drawing graphs. I enjoy the problems dealing with acceleration, velocity, and position. We did those in physics.

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12:22:00

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

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RESPONSE -->

-32t^2 / 2 + 100t + C

-16 * t^2 + 100t + C

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

C = 50

-16 * t^2 + 100(t) + 50

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12:22:34

What is the solution satisfying the given initial condition?

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RESPONSE -->

-16 * t^2 + 100(t) + 50

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12:23:47

What is the general solution to the differential equation?

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RESPONSE -->

I am not sure what the answer should be. But from my understanding of the question, my answer is -16 * t^2 + 100(t) + C

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12:31:12

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

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RESPONSE -->

v(t) = -32t+40

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

Building is 30 feet tall, so s(0) = 30 = C

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12:33:09

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

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RESPONSE -->

The balloon hits the ground when s(t) = 0

s(t) = 0 when t = 3.104

v(3.104) = -32(3.104) + 40 = -59.328 feet/sec

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12:34:54

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

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RESPONSE -->

The balloon hits the person's head when s(t) = 6

6 = -16t^2 + 40t + 30

0 = -16t^2 + 40t + 24

s(t) = 6 when t = 3

v(3) = -32(3) + 40 = -56 feet/sec

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12:35:38

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

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RESPONSE -->

v(1.5) = -8

v(3) = -56

-56 + -8 = -64

-64 / 2 = -32 feet/sec

average velocity =

change in position / change in clock time =

(s(3) s(1.5) ) / (3 sec 1.5 sec) =

(6 ft 54 ft) / (1.5 sec) =

-32 ft / sec.

Alternatively since the velocity function is linear, the average velocity is the average of the velocities at the two given clock times:

vAve = (v(1.5) + v(3) ) / 2 = (-56 ft / sec + (-8 ft / sec) ) / 2 = -32 ft / sec.

This method of averaging only works because the velocity function is linear.

The latter is the method you used; be sure you remember that it only applies when the function is quadratic, with a linear derivative.

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12:37:24

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

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RESPONSE -->

v(t) = -32 + 40

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12:40:58

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

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RESPONSE -->

I didn't really understand this section of the text. I would guess the derivative of the integral is just the original problem. But i will look over my notes and the answer to this problem and practice again later.

You guessed correctly, but be sure you understand the reasoning:

** 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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12:41:00

What is the desired derivative?

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12:42:04

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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12:42:08

Why do we use something besides x for the integrand?

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12:44:59

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

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I'm not sure what to do about the limits. But using the chain rule, the derivative of e^t^2 is 2t * e^(t^2)

Good start.

** 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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12:45:41

What is the desired derivative?

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RESPONSE -->

e ^ (t^2)

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12:46:29

How did you apply the Chain Rule to this problem?

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RESPONSE -->

z = t^2

y = e^z

z' = 2t

y' = e^t

2t * e^(t^2)

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12:46:34

Why was the Chain Rule necessary?

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