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course Mth 277
10-7-12. 8pm.
120924Assignment 10: q_a_10.3, Text Section 10.3, Problem Set 10.3, query_10.3 *
Assignment 11: q_a_10.4, Text Section 10.4, Problem Set 10.4, query_10.4
Assignment 12: q_a_10.5, Text Section 10.5, Problem Set 10.5, query_10.5
`q001. Consider the vector function `R(t) = A cos(f(t)) `i + A sin(f(t)) `j, a <= t <= b, where f(t) is some twice differentiable function of the parameter t defined on the interval [a, b].
Find the vector functions `R ' (t) and `R '' (t).
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R' = - A f ' sin(f)i + A f ' cos (f)j.
R'' = [-A sin(f) f '' - f '^2 A cos(f)]i + [f '' A cos(f) - f '^2 sin (f)]j
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Prove the following, all of which follow from an appropriate interpretation of the above derivatives:
The trace of `R(t) lies entirely on the circle of radius A centered at the origin.
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A cos (f)i + A sin(f)j.
The magnitude of this is always 1 * A, thus the vector is always of length A. The function f(t) simply becomes a number with a value of t, thus this function has no effect on the position of R(t). Nothing has been multiplied by either sin or cos to translate the graph, thus the center is at the origin. We know the values of cos and sin will produce a circle.
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The function V(t) = `R ' (t) is tangent to the circle for any value of t.
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If you take the dot product of R and R', it will always be 0. Thus, because R is the radius vector of the circle, and R' or V is always orthogonal to that radius, we can say that V is always tangent to the circle.
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The normal component of the vector A(t) = `R '' (t) is zero if and only if f '(t) = 0.
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[-A sin(f) f '' - f '^2 A cos(f)]i + [f '' A cos(f) - f '^2 sin (f)]j
Normal component of this vector is parallel to R, orthogonal to R'.
Normal component: - f '^2 A cos(f) i - f '^2 sin(f) j.
If f ' is not equal to zero, the normal component will always have a value because when sin (0) = 0, cos(0) = 1 and when cos(pi/2) = 0, sin (pi/2) = 1.
Thus, the only time the whole thing equals zero is when f ' = 0.
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The normal component of A(t) is always directed toward the origin.
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- f '^2 A cos(f) i - f '^2 sin(f) j.
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The vector
- f '^2 A cos(f) i - f '^2 sin(f) j.
can be expressed as
-f ' ^ 2 ( A cos(f) `i + sin(f) `j).
The original vector is A cos(f) `i + A sin(f) `j.
These vectors aren't multiples of one another, but I believe it's just because you left out a symbol.
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Whether f ' is positive or negative, the whole function will always be negative and therefore pointing towards the origin.
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The tangential component of A(t) is zero if and only if f ''(t) = 0.
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Tangential: - f '' A sin(f)i + f '' A cos(f)j.
For the same reasons as the normal component, when f'' not equal zero, this function always has some value. Thus it is only equal to 0 if f '' = 0.
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The tangential component of A(t) is in the direction of the unit tangent vector if and only if f ''(t) is positive, and is in the direction opposite the unit tangent vector if and only if f ''(t) is negative.
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- f '' A sin(f)i + f '' A cos(f)j.
Unit Tangent = R' / ||R'||. Thus, this is the unit vector in the direction of V.
Tangential component is parallel to V.
V = - A f ' sin(f)i + A f ' cos (f)j.
If f '' is positive, the signs of the i and j components of V and A tangential are the same, thus they run in the same direction. If f '' is negative, the signs switch and therefore A tangential runs in the opposite direction of V.
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The vector `A(t) is always perpendicular to the unit tangent vector if and only if f(t) is a linear function.
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Unit Tangent = R' / ||R'||.
V/ ||V|| = - sin (f) i + cos(f) j
A = [-A sin(f) f '' - f '^2 A cos(f)]i + [f '' A cos(f) - f '^2 sin (f)]j.
f = linear function. M x + B. f ' = M. f '' = 0.
A then becomes:
[-A sin(f) 0 - M^2 A cos(f)]i + [0 A cos(f) - M^2 sin (f)]j.
- M^2 A cos (f) i - M^2 sin(f) j.
Dot this with Unit Tangent vector and the result is 0. Thus they are perpendicular/orthogonal.
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The trace of `R(t) will be at the point (A, 0) whenever f(t) = 2 pi n, where n is an integer (note that if 2 pi n is not in the range of t, the trace will not include the point (A, 0). The trace is just the set of the points through which the particle moves, the points traced out by the tip of the changing vector `R(t); more technically it's the set of points `R(t), a <= t <= b).
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`q002. For the function of the preceding question, interpreting t as clock time, what is the expression for the speed of a particle whose position function is `R ( t )?
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Speed = magnitude of velocity.
V = R' = - A f ' sin(f)i + A f ' cos (f)j.
|| - A f ' sin(f)i + A f ' cos (f)j || = A f '.
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What is the expression for the unit tangent vector?
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R ' / ||R ' ||
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This result is general. It's not specific to the present situation.
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What is the expression for the rate at which the unit tangent vector is changing with respect to t?
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( R ' / ||R ' || ) '
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again general; we need a specific answer in terms of the given function
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Curvature is defined to be the rate of change of the unit tangent vector with respect to distance traveled on the trace.
Explain why this definition makes sense. You might think about driving around the same curve at different speeds.
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If the unit tangent vector is changing, this represents the same direction as velocity , so this is the direction of your travel. If you think about this in small increments, every time you go around a curve, your forward direction line changes. Thus, the more quickly it is changing, the more sharply you are turning. Combine this with respect to the distance traveled, if your rate of turning is greater during the same distance than in a previous scenario on a different curve, the curvature of the curve your on now must be greater than the curvature of the previous curve.
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Different paramaterizations of the same curve will typically arrive at the same point of the curve for different values of the parameter. For example the parameterization of the preceding question will arrive at the point (A, 0) whenever f(t) = 2 pi n, where n is an integer.
The faster you travel along the trace the faster your direction of motion changes. At a given point on the trace, why would you expect `T ' (t) / speed to be independent of the paramaterization?
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Because this defines the curvature of the trace and no matter the parameterization, it is still defining the same curve. Thus, no change in curvature, so T'(t)/speed will be independent of the paramaterization.
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How should the rate of change of `T with respect to t be related to the rate of change of distance with respect to t and to the curvature?
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dT/dt = dT/ds * ds/dt.
||dT/ds || is curvature.
dT/dt is rate of change of T with respect to time.
ds/dt is rate of change of distance with respect to time.
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In terms of `R(t) and its derivatives, what is the expression for curvature?
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curvature = || T ' ||/ ||R ' ||
Where T ' = ( R' / ||R' ||) '.
Thus:
|| ( R' / ||R' ||) ' || / || R' ||
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`q003. What is the trace of the vector function `R(t) = t `i + sqrt(1 - t^2) `j?
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x = t.
y = 1 - x^2.
When x = 0, y = 1.
when y = 0, x = +- 1.
From y = 1- x^2, we see the trace in the xy plane is a parabola.
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Show the following for this vector function:
The magnitude of `R(t) is constant.
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||R|| = sqrt( t^2 + (sqrt(1 - t^2))^2 )
||R|| = sqrt( t^2 + 1 - t^2)
||R|| = sqrt(1) = 1.
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The first and second derivatives of `R(t) exist only for -1 < t < 1.
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`R(t) = t `i + sqrt(1 - t^2) `j.
R' = i + -2t/[ 2* sqrt(1-t^2)] j.
Anything outside -1 < t < 1 would create a negative inside the square root in the denominator of the j component, which produces imaginary numbers.
R'' = - 1/ (1-t^2)^(3/2) j.
For the same reasons, t must lie between -1 and 1.
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The rate of change of arc distance with respect to t is not constant.
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`R(t) = t `i + sqrt(1 - t^2) `j.
int( sqrt( dx/dt ^2 + dy/dt^2 ), 0 to t).
int ( sqrt( 1 + t^2/ (1-t^2) ).
this comes out to be:
sqrt( 1/ (1-t^2) ) * sqrt( 1-t^2) * arcsin (t) (by way of technology).
If you plug values in for t, it becomes evident the value changes as t does.
If the function changes, the derivative must also change.
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The rate of change of arc distance with respect to t approaches infinity as | t | approaches 1.
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The rate of change of the arc distance (using technology) is:
sqrt( 1/ (1-t^2) ).
As t approaches 1, the difference between 1 and t^2 becomes very small and approaches 0. Thus, 1/ (an increasingly smaller number) produces a huge number. Taking the square root of this doesn't change the fact that it is a huge number, and will approach infinity as t grows close to 1.
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`R ' (t) is always perpendicular to `R(t).
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`R(t) = t `i + sqrt(1 - t^2) `j.
R' = i + -t/[ sqrt(1-t^2)] j.
(t * 1 ) + sqrt(1-t^2) * -t/sqrt(1-t^2).
t - t = 0.
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`R '' (t) is perpendicular to `R(t) at only one point.
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`R(t) = t `i + sqrt(1 - t^2) `j.
R'' = - 1/ (1-t^2)^(3/2) j.
Dot product gives:
-1/ (1- t^2).
This can't be equal to zero... I obviously messed up somewhere.
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The tangential component of `R '' (t) is in the direction of the unit tangent vector for every point in the second quadrant, and opposite the direction of the unit tangent vector for every point in the first quadrant.
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y = 1 - x^2.
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The curvature of the trace is always 1.
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If the magnitude is always a constant 1, and curvature can be represented as 1/r, then it follows that the curvature is always 1/1 or 1.
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`q004. The vector function `R(t) = tau `i + sqrt(1 - `tau^2) `j, with tau = (t - 1) / 2 - greatestInteger( (t - 1) / 2), has the same trace and the same properties as the function in the preceding question. [greatestInteger(x) is the largest integer which is not greater than x; so for example greatestInteger(2.5) = 2, greatestInteger(12) = 12, greatestInteger(-1.5) = -2, greatestInteger(pi) = 3, greatestInteger(-pi) = -4.]
Describe the motion of the point defined by `R(t) for -2 <= t <= 2.
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t = -2. R = .5 i + sqrt(.75) j.
t = -1. R = 0 i + j.
t= 0. R = .5 i + sqrt(.75) j.
t = 1. R = 0 i + j.
t = 2. R = .5 i + j.
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You've given a few points, but you haven't described the motion.
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What is the average rate of change of `R(t) with respect to t between t = .9 and t = 1.1?
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t = 1.1. R = .05 i + .99 j.
t = .9. R = .95 i + .31 j.
(y2 - y1)/(x2-x1).
(.99 - .31)/(.05 - .95) = -.75.
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`R(t) is a vector.
`R(1.1) - `R(0.9) = .9 `i - .68 `j, so
`d(`R) / `dt = (.9 `i - .68 `j) / (1.1 - 0.9) = 4.5 `i - 3.4 `j.
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What is the average rate of change of `R(t) with respect to t between t = 1 - .1^n and t = 1 + .1^n, for positive integer n?
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tau for t = 1- .1^n is 1 - .1^n/2.
tau for t = 1+ .1^n is .1^n/2 .
d y = sqrt( 1 - (.1^n/2 )^2 ) - sqrt ( 1 - (1 - .1^n/2)^2 ).
As n increases, this expression will approach 1.
d x = .1^n/2 - 1 + .1^n/2.
d x = .1^n -1. This approaches -1.
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The average rate is (change in vector) / (change in t) = (vector change) / (change in t), which is a vector.
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What is the limiting value of the average rate of change of `R(t) with respect to t between t = 1 - .1^n and t = 1 + .1^n, for positive integer n?
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Well... if n approaches infinity, the slope approaches -1?
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What is wrong with the last question?
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I don't know.
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The function is not defined for a point between the two values of t (specifically it's not defined for t = 1). So, with no information for part of the interval, you can't take an average of anything over this interval.
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Think about continuity and discontinuity.
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You've done a nice job on most of these questions.
Naturally there are notes for you to consider.
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