Assignment 2

course MTH 174

ƫ𡦠sWկassignment #002

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Physics II

10-15-2007

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01:39:13

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

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RESPONSE -->

f(t) = -16t + 100t + C

F(t) = 50 when t = 0

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

C = 50

f(t) = -16t + 100t + C

Should be f(t) = -16 t^2 + 100 t + C.

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01:39:41

What is the solution satisfying the given initial condition?

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RESPONSE -->

C = 50

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01:40:31

What is the general solution to the differential equation?

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RESPONSE -->

f(t) = -16t + 100t + 50

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02:15:25

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

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RESPONSE -->

(a) h(t) = -16t^2 + 40t + 30

(b) v(t) = -32t + 40

v(1.5) = -8

v(3) = -56

Average Velocity = (8 + 56) / 2 = 32 ft/sec.

(c) 6 = -16t^2 + 40t + 30

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

t = 3

v(3) = 56 ft/sec. when it strikes the person on the top of the head.

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02:19:39

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

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RESPONSE -->

59.33 ft/sec

** 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. If the accurate solution is used we find that the ball strikes the ground at precisely 60 ft / sec.

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

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02:19:58

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

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RESPONSE -->

56 ft/sec

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02:20:23

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

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RESPONSE -->

32 ft/sec

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02:20:41

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

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RESPONSE -->

v(t) = -32 +40

Good. More detail:

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.

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11:56:36

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

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RESPONSE -->

(a) 6 seconds

(b) left and right sums? I know what would be and over estimate and what would be an underestimate.

This should give the exact amount: 27.5+22.5+17.5+12.5+7.5+2.5 = 90 ft. traveled

(c) Half the base times the height of the graph equals the area underneath. 3 sec * 30 ft / sec = 90 ft.

(d) v(x) = -5x + 30

intergral from 0 to 6 sec. of -5x + 30 = F(b) - F(a) = -5/2(6)^2 - -5/2(0)^2 = -90, a distance of 90 ft.

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11:57:33

What is the desired derivative?

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RESPONSE -->

V(x) = -5/2 x^2

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12:19:15

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 knew from the table exactly when the car came to rest.

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12:23:45

Why do we use something besides x for the integrand?

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RESPONSE -->

x was replaced by t for time. The integrand was v for velocity.

There's an apparent discrepancy between text and query. I don't have a copy of the text at this location and can't tell whether your solution matches the problem you are solving, but everything you say makes sense (except that halfway between left and right-hand sums doesn't generally give the exact result--functions don't usually have that sort of linearity).

Here is a solution to the problem given here. Be sure you understand it and let me know if you don'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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22:22:23

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

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RESPONSE -->

-(e^(cos^(2)(x))sin(x)

If we were finding (int(e^(t^2),t, x, 0) the answer would just be e^(x^2) by the Second Fundamental Theorem (along with the reversal of integration limits and therefore sign).

However the lower limit on the integral is cos(x). This makes the expression int(e^(t^2),t,cos(x), 3) a composite of f(z) = (int(e^(t^2),t, z, 3) and g(x) = cos(x).

Be sure you see that the composite f(g(x)) = f(cos(x)) would be the original expression int(e^(t^2),t,cos(x),0).

g'(x) = -sin(x), and by the Second Fundamental Theorem f'(z) = e^(z^2) (again the negative is because of the reversal of integration limits).

The derivative is therefore

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

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22:23:07

What is the desired derivative?

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RESPONSE -->

-(e^(cos^(2)(x))sin(x)

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22:24:24

10-16-2007 22:24:24

How did you apply the Chain Rule to this problem?

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NOTES -------> The derivative of the whole thing times the derivative of what's inside.

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22:25:56

Why was the Chain Rule necessary?

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RESPONSE -->

It would have been very extensive work otherwise.

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Good work. See my notes and let me know if you have questions.