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course Mth 277
Ignore that first submission without those access codes11/22 12am
query_10_5
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Question: Find the tangential and normal components of an object's acceleration which has the position vector R(t) = <3/5 cos t, 4/5(1+sin t), cos t>.
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Your solution:
T = R ' / || R ' ||
N = T ' / || T ' ||
This obviously is a very large derivative. I will write it all out on paper
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Given Solution:
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Self-critique (if necessary):
I can only get it so far.
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Question: If V(0) = <5,-2,4> and A(0) = <1,3,-9>, what is A_T and A_N at t = 0?
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Your solution:
A_t = V dot A / || V ||
A_n = V cross A / || V ||
I used the v(0) and a(0) numbers to get this data. But for some reason I think i should use the ""final v and a"", but I don't know how to get to this.
But I""m pretty sure I understand the concept
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@& A_T is the component of a in the direction of V, while A_N is the component of a in a direction perpendicular to V.
So A_T is the projection of A onto V, and you can get A_N by subtracting A_T from A.*@
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Question: An object moves with a constant angular velocity omega around the circle x^2 + y^2 = r^2 in the xy-plane.
Find a parameterization for the circle.
Compute the tangential and normal acceleration for the object.
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Your solution:
well A_n = omega^2 ( r )
and A_t = omega (r)
where r is just sqrt(x^2 + y^2)
no omega was given.
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Given Solution:
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@& You haven't parameterized the circle.
One possible parameterization is
What functions x(t) and y(t) parameterize the circle for the given conditions?*@
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Question:
Consider the vector function R(t) = <3 sin t, 4t, 3 cos t>.
Evaluate V(t) = R'(t), N(t), and A(t) = R''(t) when t = 1.
Find the vector projection of A(1) onto V(1). Denote this proj_V(1) (A(1)).
Find the vector projection of A(1) onto N(1). Denote this proj_N(1) (A(1)).
What is the sum of proj_V(1) (A(1)) and proj_N(1) (A(1)).
How does proj_V(1) (A(1)) relate to A_T when t = 1.
How does proj_N(1) (A(1)) relate to A_N when t = 1.
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Your solution:
I'm in radian form in the calculator. which seems appropriate.
N(1) i'm getting i + j - k
A(1) is just R''(t) = -2.5i + j - 1.6k again in radian mode
proj_N(1) (A(1)) just cross A(1) onto V(1) = -0.6i - 4.1j - 3.5k
now proj_V(1)(A(1)) = 0.9i +8.2j + 4.1k
the sum is 0.3i + 4.1j -0.6k
proj v onto a related to A_t because the velocity components tangential to the curve are slowing down because acceleration is negative in that direction tangential to the curve.
proj N to A related to A_n because the normal path of acceleration is influenced by the amount of acceleration back to the direction its actually accelerating. if it is strong, then the curve will be tight, typically, the greater the
normal acceleration, the less the tangential.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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@& R ' (1) is roughly 1.5 `i + 4 `j -3.5 `k.
R '' ( 1) is roughly -2.5 `i - 1.5 `k, which pretty much agrees with your result except that the `j component should be zero.
So A(1) has a magnitude of 3, its dot product with V(1) has magnitude about 1 and the angle of A(1) with V(1) is pretty close to 90 degrees. Its projection onto V(1) has magnitude significantly less than 3. Your corresponding projection vector has magnitude close to 10, which is not possible. A projection of a vector cannot be of magnitude greater than the vector itself.
Your calculated normal component also has magnitude greater than A(1), which is not possible.
So you're on probably the right track with the vectors themselves, but your projections aren't right. *@
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Question: Let B = T X N when T and N are the unit tangent and normal vectors to a curve C with position vector R. Show that dB/ds = T X (dN/ds).
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Your solution:
well I've seen the hints here in the book and as far as I can sypher is 2B = dN/ds cross N. for T=dN/ds
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Given Solution:
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@& What is the expression for the derivative of T X N with respect to s? You need this expression to get started on this problem.*@
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