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course Mth 277
11:50 pm 9/25/12
120919Assignment 10: q_a_10.3, Text Section 10.3, Problem Set 10.3, query_10.3 *
`q001. There are some values of t for which you wouldn't bother trying to plot the vector function `F(t) = sin(t) `i + t^2 `j + sqrt(t+3) `k . What are these values?
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A number less than -3. This is outside the domain of the function.
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`q002. A projectile is given an initial velocity in the direction 10 degrees below horizontal, and it lands on a level floor at a point 1 meter lower and 25 centimeters from the point directly below the initial point.
Model the acceleration, velocity and position of this projectile as vector functions and see if you can determine the initial velocity of the projectile and the time required for it to fall.
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If you let the point from where the projectile was fired be your datum, then that point is (0,0), the point where it lands is (25 cm, -100 cm).
Let Va be the projectile velocity.
sf = s0 + v t + 1/2 at^2.
Looking at this in the x direction:
No acceleration in the x direction. a = 0.
v = Va cos(10).
sf (final destination) = 25.
s0 = 0.
25 = vA cos (10) t.
vA = 25/ (t* cos(10))
Y Direction:
s0 = 0
sf= -100.
v = -vA sin(10) since its in the downward direction.
a = -9.81 m/s^2 or 981 cm/s^2.
-100 = 0 - vA sin(10) t - 490.5 t^2. Next plug in vA obtained from x direction equation.
-100 = -25/(t*cos(10)) * sin(10) t - 490.5 t^2. t's cancel, sin/cos becomes tan.
-100 = -25 tan(10) - 490 t^2.
Solve for t
490 t^2 = -25 tan(10) + 100.
t^2 = (-25 tan(10) + 100)/490.
t = .44 seconds.
Plug this back into vA =25/ (t* cos(10)).
vA = 57.7 cm/s.
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Your solution doesn't treat the position function as a single vector function. You ultimately get the same result but not the same conceptual framework.
Assuming the projectile to start from the origin (this is an arbitrary assumption; we could have started as well at (0, 100 cm), but the present assumption is equivalent to yours).
`R(t) = v0 cos(10 deg) * t `i + (v0 sin(350 deg) * t - 1/2 g t^2) `j
The given condition is that for some value of t, `R(t) = 25 `i - 100 `j.
So we want to solve the equation
v0 cos(10 deg) * t `i + (v0 sin(350 deg) * t - 1/2 g t^2) `j = 25 `i - 100 `j.
This leads to the simultaneous equations
v0 cos(10 deg) * t = 25
v0 sin(350 deg) * t - 1/2 g t^2 = -100.
There are many ways to solve this system, which is of course identical to the system you solved.
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`q003. Prove whether the derivative of `F_1 `X `F_2 is `F_1 ' `X `F_2 + `F_1 `X `F_2 ' , where `F_1 and `F_2 are vector functions of t, the derivatives are with respect to t and `X indicates the cross product.
Don't type in all the details of your proof, which would take an unreasonable amount of time, but do give a reasonable explanation of how you proceeded and what you found.
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X where F_1 is the first and F_2 is the second.
x1 g1 i - x2 g2 j + x3 g3 k.
Derive this.
(x1 g1' + g1 x1') i + (-x2 g2' - g2 x2') j + (x3 g3' + g3 x3')k
F_1' = < x1', x2', x3'>
cross with F_2. x1' g1 i - x2' g2 j + x3' g3 k.
F_2' =
F_1 cross with F_2'. x1 g1' i - x2 g2' j + x3 g3' k.
Add cross products together to get:
(x1' g1 + x1 g1' )i + (- x2' g2 - x2 g2')j + (x3' g3 + x3 g3')k.
Same as above.
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This looks like a dot product but with `i, `j and `k vectors where they shouldn't be.
The cross product would be
(x2 g3 - x3 g2) `i + (x1 ge - x3 g1) `j + (x1 g2 - x2 g1) `k.
Still you're on the right track.
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&#This looks good. See my notes. Let me know if you have any questions. &#