Asignment 2

course Mth 174

bzassignment #002

د`SB

Physics II

06-03-2009

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15:46:11

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

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RESPONSE -->

ds/dt = -32 t + 100

s= -4.9t^2 + 100t + K

50 = -4.9(0)^2 + 100(0) + K

50 = K

s= -4.9t^2 + 100t + 50

Right procedure, but 4.9 doesn't appear anywhere in the solution to the problem as stated here.

However if expressed in metric units, you would get -4.9 t^2. However 100 and 50 would also change due to the change in units.

For the problem as stated:

s' = 100 - 32t.

Integrating with respect to t we obtain

s= 100t - 16t^2 + C.

Since s = 50 when t = 0 we have

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

which we easily solve to obtain

50 =C.

this into the expression for s(t) we have

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

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15:46:34

What is the solution satisfying the given initial condition?

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RESPONSE -->

Yes because I plugged in 0 for t when s=50.

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15:47:03

What is the general solution to the differential equation?

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RESPONSE -->

s= -4.9t^2 + 100t + K

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15:56:03

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

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RESPONSE -->

v(t) = -32t+40

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

30= -16(0)2 + 40(0) + K

30=K

s(t)= -16t^2 + 40t + 30

0= -16t^2 + 40t + 30

simplify:

(4t + 6) (4t + 4)

Right idea, but this factorization doesn't work. This product would give you 16 t^2 + 40 t + 24, not -16 t^2 + 40 t + 30.

t=-1.5 sec

t=-1 sec

I'm not sure which time I should use in the velocity equation but I would plug in t into v(t) = -32t+40 to find the speed when it strikes the ground. How do I know which time to use?

Since v(t) = s ' (t), it follows that the antiderivative of the v(t) function is the s(t) function so we haves

s(t) = -16 t^2 + 40 t + c.

Since the building is 30 ft high we 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.

The interpretation of this result is that when it strikes the ground the balloon is moving downward at 60 feet / second.

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15:57:50

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

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RESPONSE -->

v(t) = -32t+40

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

30= -16(0)2 + 40(0) + K

30=K

s(t)= -16t^2 + 40t + 30

0= -16t^2 + 40t + 30

simplify:

(4t + 6) (4t + 4)

t=-1.5 sec

t=-1

I would plug in t into v(t) = -32t+24 to find the speed when it strikes this person. I use 24 instead of 30 because you must subtract 6ft from 30ft. Again, I don't know which time to use but I understand how to plug it in.

** You have to determine the clock time t when the balloon's altitude is 6 feet. You have the s(t) function. So at what time is the altitude 6 ft?

To answer the question we solve the equation

s(t) = 6, i.e.

-16t^2+40t+30=6. This is a quadratic equation. We rearrange to get

-16t^2+40t+24=0; dividing by -16 we have

t^2-5/2t-3/2=0.

We can solve using the quadratic formula or by factorization, obtaining

t=3 or -.5

When t=3, we have

v(3)=-32*3+40=-56.

Thus the velocity at the 6 ft height is 56 ft/sec downward.

**

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15:59:36

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

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RESPONSE -->

v(t) = -32t+40

v(t)= -32(1.5) + 40

v= -8

v(t)=-32(3) + 40

v(t)=-56

(-8 + -56) / 2 = -32 = average velocity

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16:00:00

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

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RESPONSE -->

v(t) = -32t+40

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16:03:52

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

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RESPONSE -->

derivative of (int(ln(t)), t, x, 1)

ln(1) - ln (x) = -lnx

You end up with the right answer, but ln(1) is not part of the derivative. Be sure you understand:

** 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 find an antiderivative.

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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16:03:57

What is the desired derivative?

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RESPONSE -->

derivative of (int(ln(t)), t, x, 1)

ln(1) - ln (x) = -lnx

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16:04:48

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 still plugged in the upper limit first and subtracted it from the lower limit plugged in to the equation.

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16:05:33

Why do we use something besides x for the integrand?

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RESPONSE -->

We use something besiedes x in order to be able to substitue both the upper and lower limit, rather than only the x value.

** 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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16:13:46

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

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RESPONSE -->

derivative of (int(e^-(t^2),t, 0,x^3)

I don't understand what the last part of the problem means where you have t, 0,x^3. This problem in the book has an upper limit of 3 and a lower limit of cos(x) and this is what I got:

e^(3^2) - e^(cos(x)^2)

If you mean that 0 and x^3 are the limits then,

e^((x^3)^2) - e^(0^2)

(int(e^-(t^2),t, 0,x^3) means the integral of e^(-t^2) with respect to t over the interval from 0 to x^3.

** 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 understand how the composite f(g(x)) = f(x^3) would be the original expression int(e^(-t^2),t,0,x^3).

Now we apply the chain rule:

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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16:13:53

What is the desired derivative?

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RESPONSE -->

derivative of (int(e^-(t^2),t, 0,x^3)

I don't understand what the last part of the problem means where you have t, 0,x^3. This problem in the book has an upper limit of 3 and a lower limit of cos(x) and this is what I got:

e^(3^2) - e^(cos(x)^2)

If you mean that 0 and x^3 are the limits then,

e^((x^3)^2) - e^(0^2)

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16:18:00

How did you apply the Chain Rule to this problem?

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RESPONSE -->

The chain rule could be applied by saying f(x)=e and t^2=g(x) and f ' (g(x)) * g (x) then,

e * t^2 * 2t

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16:18:39

Why was the Chain Rule necessary?

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RESPONSE -->

I'm not sure but I don't think it is necessary when the function includes e

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