Asst 2

course Mth 174

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Physics II

06-15-2007

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02:50:12

What is the solution satisfying the given initial condition?

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RESPONSE -->

ds/dt = -32t +100

s = -16t^2 + 100t + c

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

c = 50

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02:50:29

What is the general solution to the differential equation?

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RESPONSE -->

s = -16t^2 + 100t + c

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02:57:28

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

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RESPONSE -->

V(t) = -32t + 40

S(t) = -32

S(0) = -32 ft/sec

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03:02:33

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

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RESPONSE -->

Intgral from 30 to 6 v(t) = -32t + 40

-16t^2 + 40t + 30 from 30 to 6

[-16(30)^2 + 40(30) + 30] - [-16(6)^2 + 40(6) + 30] = -12558

This can't be correct but I'm confused.

If you integrate v(t) a definite integral gives you the change in position, because you are summing up v(t) * `dt intervals (v(t) * `dt is approximate displacement).

To get the speed you have to find when the balloon reaches that position, the plug that clock time into the velocity function.

You have to determine the clock time t when the balloon's altitude is 6 feet. You have the s(t) function.

At what time is the altitude 6 ft? What is the velocity at that clock time?

You should try to answer this yourself before looking at the solution provided below.

** -16t^2+40t+30=6
-16t^2+40t+24=0
t^2-5/2t-3/2=0.

Quadratic formula or factorization give
t=3 or -.5

When t=3,
V(3)=-32*3+40=-56.

You can work this out more precisely; all calculations here done mentally. **

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03:03:47

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

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RESPONSE -->

-32 (1.5) + 40 = -8

-32 (3) + 40 = -56

-56+-8 = -64/2 = -32

Among other things your statement says that -56+-8 = -32. This is clearly not true.

Use = signs to indicate equality, not train of thought.

-56+-8 = -64
-64/2 = -32

or just
(-56 + 8) / 2 = -32.

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03:04:14

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

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RESPONSE -->

v(t) = -32t + 40 ft/sec

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03:06:01

What is the desired derivative?

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RESPONSE -->

ln (1) - ln (x) = 0 - ln (x) = -ln(x)

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03:09:02

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 just found the integral and then placed a negative in front because the int(f(t)) (a, b) = -int(f(t)) (b,a)

** 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 therefore g'(x) f'(g(x))= 3x^2 * e^-( (x^3)^2 ). This one takes a bit of thinking.**

This is easily adapted to the current problem:

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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03:09:37

Why do we use something besides x for the integrand?

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RESPONSE -->

the lower limit needs to be constant not a variable.

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03:13:27

What is the desired derivative?

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RESPONSE -->

e^9-e^((cosx)^2)

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03:14:04

How did you apply the Chain Rule to this problem?

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RESPONSE -->

There should be a 2t in front of the e^(t^2)

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03:14:17

Why was the Chain Rule necessary?

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RESPONSE -->

you must find the derivative of the exponent.

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Asst 2

If you integrate v(t) a definite integral gives you the change in position, because you are summing up v(t) * `dt intervals (v(t) * `dt is approximate displacement). To get the speed you have to find when the balloon reaches that position, the plug that clock time into the velocity function.

Among other things your statement says that -56+-8 = -32. This is clearly not true. Use = signs to indicate equality, not train of thought. -56+-8 = -64 -64/2 = -32 or just (-56 + 8) / 2 = -32.

Your work isn't bad, but see my notes and be sure you are seeing all the details in these problems. Let me know if you have specific questions.

If you integrate v(t) a definite integral gives you the change in position, because you are summing up v(t) * `dt intervals (v(t) * `dt is approximate displacement).

To get the speed you have to find when the balloon reaches that position, the plug that clock time into the velocity function.

Among other things your statement says that -56+-8 = -32. This is clearly not true.

Use = signs to indicate equality, not train of thought.

-56+-8 = -64
-64/2 = -32

or just
(-56 + 8) / 2 = -32.