course Mth 174
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Physics II
02-04-2009
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11:37:15
query 6.5 #8 Galileo: time for unif accel object to traverse dist is same as if vel was ave of init and final; put into symbols and show why true
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RESPONSE -->
This statement is true because if you take the area under the curve for the graph this statement represents, it would be the same under any given points, since the gravitational pull is constant and acceeration of the body is constant.
s= - gt^2/2 + v(0)t + s(0)
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11:39:07
how can you symbolically represent the give statement?
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RESPONSE -->
Acceleration = dv/dt + -g so
v = -gt + C
v= -gt + v(0) = ds/dt
s= - gt^2/2 + v(0)t + C
s= - gt^2/2 + v(0)t + s(0)
** Using s for the distance fallen we can translate Galileo's statement as follows:
t = s / [ (vf + v0)/2 ].
A ball dropped from rest will have position function s = .5 a t^2; since a = 32 ft/s^2 we have s = 16 t^2.
We assume that the time of fall is t. Then using Galilieo's assumption we show that they are consistent with the result we get from s = 16 t^2.
If an object is dropped from rest and falls for time t it will reach velocity vf = a t = 32 t. So the average of its initial and final velocities will be (vf + v0) / 2 = (32 t + 0) / 2 = 16 t.
The distance fallen is s = 16 t^2.
The time to fall distance s = 16 t^2 at average velocity 16 t is s / t = 16 t^2 / (16 t) = t, which agrees with the time the object was allowed to fall.
A numerical example for a given s:
When s = 100, then, t is the positive solution to 100 = 16 t^2. This solution is t = 2.5.
The velocity function is v = a t. We have to find v when s = 100. When s = 100, we have t = 2.5, as just seen. So v = a t = 32 * 2.5 = 80, representing 80 ft/sec.
Now we know that s = 100, vf = 80 and v0=0. Substituting into t = s / [ (vf + v0)/2 ] we get t = 100 / [ (80 + 0) / 2 ] = 100 / 40 = 2.5.
This agrees with the t we got using s = .5 a t^2. **
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11:40:24
How can we show that the statement is true?
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RESPONSE -->
Acceleration = dv/dt + -g so
v = -gt + C
v= -gt + v(0) = ds/dt
s= - gt^2/2 + v(0)t + C
s= - gt^2/2 + v(0)t + s(0)
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11:41:33
How can we use a graph to show that the statement is true?
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RESPONSE -->
The graph of this statement would show the constant velocity under constant gravitational pull within a given period of time.
** Using s for the distance fallen we can translate Galileo's statement as follows:
t = s / [ (vf + v0)/2 ].
An object accelerating for time t, starting with initial velocity v0 at t = 0, has velocity function
v = v0 + a * t
and position function
s = .5 a t^2 + v0 t,
assuming that s = 0 at t = 0.
Both of these results are easily obtained, for the given initial condition that initial velocity is zero, by a straightforward integration of the acceleration function (to get the velocity function) then the velocity function (to get the position function).
For given displacement s we can solve the position equation for t. The equation is rearranged to the form of a standard quadratic:
.5 a t^2 + v0 t – s = 0, with solutions
t = (-v0 +- sqrt(v0^2 + 2 a s) ) / a ).
Substituting this into the velocity function we obtain the final velocity:
Final velocity
= v0 + a t = v0 + a * (-v0 +- sqrt(v0^2 + 2 a s) / a)
= +- sqrt(v0^2 + 2 a s) .
The average of the initial and final velocities is therefore
(initial vel + final vel) / 2 = (v0 +- sqrt(v0^2 + 2 a s)) / 2.
Traveling at this velocity for time t = (-v0 +- sqrt(v0^2 + 2 a s) ) / a ) the displacement will be
Displacement = velocity * time
= ( v0 +- sqrt(v0^2 + 2 a s)) (-v0 +- sqrt(v0^2 + 2 a s) ) / a ) / 2
The numerator is the product of the sum and the difference of two quantities and simplifies to 2 a s. The denominator is 2 a so the expression simplifies to just s.
This confirms that the distance traveled is the same as the distance that would be traveled at the average of the initial and final velocities.
Alternatively:
For uniform acceleration the velocity function can be expressed as
v0 + a t.
Integrating this function between clock times t1 and t2 we obtain 1/2 a (t2^2 – t1^2) + v0 ( t2 – t1).
Dividing the integral by the length of the interval we get the average value of the function—i.e., the average velocity. The length of the interval is t2 – t1 so
Integral / interval = (1/2 a (t2^2 – t1^2) + v0 ( t2 – t1)) / (t2 – t1) = 1/2 a ( t2 + t1) + v0.
The initial value of the velocity on the interval is v0 + a t1, the final velocity v0 + a t2, so the average of initial and final velocities is
Ave of init and final velocity = (v0 + a t1 + v0 + a t2) / 2 = v0 + 1/2 a ( t1 + t2).
Thus the average of the initial and final velocities is equal to the average velocity on the interval, and the result is proven.
Let t be the time of fall. Then initial velocity is 0 and final velocity is a * t, so the average of initial and final velocities is
Ave of init and final vel = (0 + a * t) / 2 = ½ a t.
Then using Galilieo's assumption we show that they are consistent with the result we get from s = 16 t^2.
If an object is dropped from rest and falls for time t it will reach velocity vf = a t = 32 t. So the average of its initial and final velocities will be (vf + v0) / 2 = (32 t + 0) / 2 = 16 t.
The distance fallen is s = 16 t^2.
The time to fall distance s = 16 t^2 at average velocity 16 t is s / t = 16 t^2 / (16 t) = t, which agrees with the time the object was allowed to fall.
A numerical example for a given s:
When s = 100, then, t is the positive solution to 100 = 16 t^2. This solution is t = 2.5.
The velocity function is v = a t. We have to find v when s = 100. When s = 100, we have t = 2.5, as just seen. So v = a t = 32 * 2.5 = 80, representing 80 ft/sec.
Now we know that s = 100, vf = 80 and v0=0. Substituting into t = s / [ (vf + v0)/2 ] we get t = 100 / [ (80 + 0) / 2 ] = 100 / 40 = 2.5.
This agrees with the t we got using s = .5 a t^2. **
The average value of any linear function over an interval is equal to the average of its initial and final values over the interval.
In a nutshell, because the v vs. t graph is linear, the average velocity is equal to the average of the initial and final velocities.
Since
time of fall = displacement / average velocity,
it follows that
time of fall = displacement / (ave of initial and final vel).
This latter expression is just the time that would be required to fall at a constant velocity which is equal to the average of initial and final velocities.
More rigorously:
The graph is linear, so the area beneath the graph is the area of a triangle.
The base of the triangle is the time of fall, and its altitude is the final velocity.
By a simple construction we know that the area of the triangle is equal to the area of a rectangle whose length is equal to the base of the triangle, and whose width is equal to half the altitude of the triangle.
It follows that the area of the triangle is equal to half the final velocity multiplied by the time interval.
Since the initial velocity is 0, the average of initial and final velocities is half the final velocity.
So the area of the triangle is the product of the time of fall and the average of initial and final velocities
area beneath graph = time of fall * ave of init and final vel
The area beneath the graph is equal to the integral of the velocity function over the time interval, which we know is equal to the displacement. So we have
displacement = time of fall * ave of init and final vel, so that
time of fall = displacement / ave of init and final vel.
This leads to the same conclusion as above.
Also, more symbolically:
A graph of v vs. t graph is linear. Over any time interval t = t1 to t = t1 + `dt, displacement is represented by the area of the corresponding trapezoid, which is d = (v1 + v2) / 2 * `dt.
Solving for `dt we again obtain `dt = d / [ (v1+v2)/2 ].
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12:06:55
query problem 7.1.22 (3d edition #18) integral of `sqrt(cos(3t) ) * sin(3t)
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RESPONSE -->
integral of sqrt (cos(3t)) * sin(3t) =
w = cos(3t)
dw= -sin(3t)
integral of sqrt(w) *dw =
integral of (w^3\2)/(3/2) *dw =
integral of (cos(3t)^3/2)/3/2 * -sin(3t)
** TYPICAL INCORRECT SOLUTION: (-2/3) (cos3t)^(3/2)
INSTRUCTOR COMMENT: The derivative of cos(3t) is -3 sin(3t), wo the derivative of -2/3 (cos(3t))^(-3/2) is 3 sqrt(cos(3t) sin(3t). This solution is not consistent with the Chain Rule.
To perform the integral use substitution. Very often the first substitution you want to try involves the inner function of a composite, and that is the case here. Cos(3t) is the inner function of the composite sqrt(cos(3t)), so we try using this as our substitution:
w = cos (3t) dw = -3 sin (3t)
so that sin(3t) = -dw / 3.
Thus our expression becomes
w^(1/2) * (-dw / 3).
The integral of -1/3 w^(3/2) with respect to w is -1/3 * (2/3) w^(3/2), treating w^(1/2) as a power function.
This simplifies to
-2/9 w^(3/2) or
-2/9 * (cos(3t))^(3/2).
The general antiderivative is
-2/9 * (cos(3t))^(3/2) + c,
where c is an arbitrary constant.**
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12:07:01
what did you get for the integral and how did you reason out your result?
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RESPONSE -->
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12:13:48
query problem 7.1.20 (3d edition #21) antiderivative of x^2 e^(x^3+1)
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RESPONSE -->
integral of x^2 e^(x^3 + 1) =
w= x^3 +1
dw= 3x^2
xdx= 1/3 ?
integral of x^2 e^w *dw
integral of (x^3/3) e^(x^3 + 1) * 3x^2
Try substituting for the inner function of the composite:
u = x^3 + 1 yields
du = 3 x^2 dx, so
x^2 dx = 1/3 du.
This makes the integrand 1/3 e^u du.
The antiderivative is 1/3 e^u + c, or 1/3 e^(x^3+1) + c.
If we take the derivative of this function we do indeed obtain our original integrand.
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In both the second and third problems you essentially assumed that the integral of a product function is equal to the product of the integrals, confusing
integral (f * g) with
integral (f) * integral(g)
(specifically, you integrated cos(t) separately from the rest of the integral in the second problem, and x^2 separately in the third problem).
This cannot lead to a correct solution. The results you get cannot lead to an expression whose derivative is equal to the original integrand.
For example the derivative of your expression ((cos(3t))^3/2)/3/2 * -sin(3t) is found using the product and chain rules to be something like
3 sin(3t) sin(t) - 2 cos(3t))^(3/2) cos(3t), not the original integrand
`sqrt(cos(3t) ) * sin(3t).
More generally we see that this simply can't work by comparing the derivative of integral (f) * integral(g) with the original integrand f * g:
(integral (f) * integral(g)) = ( integral (f) ) ' * integral(g) + integral (f) * (integral(g)) ' = f * integral(g) + g * integral(f),
which is much different than the original integrand f * g.