af 277

' title: Chp 10 Query 5

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course Mth 277

Very sorry i'm not able to figure out more of this, but with your assistance, I will hopefully know how to start it and re-submit the form.submitted 10/28 5

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:

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Given Solution:

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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:

I gave all the numbers a value of i, j, and k. anyways A_t is (v dot a) / || v || and A_n is (V X A) / || V ||

for A_t I got ( i +3j -9k ) / sqrt(45)

@& v dot a = 5*1 - 2 * 3 + 4 * (-9) = -37, so A_T = -37 / sqrt(45).

This is the scalar component of the acceleration.

The vector component is `v dot `a / || v || * `v / || v ||, which would come out to

`a_T = -37 * (5 `i - 2 `j + 4 `k)) / 45.

*@

for A_n I got (20i -98j +68k) / (sqrt(121770)

@& The denominator should be || v ||, not || `v X `a ||.*@

for me this was difficult because you said at t=0. well I have no idea where to actually put t into the equation and when to set it to zero. else the values i'm sure would be smaller for A_n

could you start the equation when to sub t in there?

@& The vectors `v(0) and `a(0) are the vectors `v(t) and `a(t) at t = 0. So if your results are correct, they will be the t = 0 results.*@

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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:

going from the point (1,2) and (4,5)

lets say (x,y) = (1,2) and (4,5) for 0<=t<=1

so (x,y) = (1,2) + t*(4,5)

x= 1 + 4t

y= 2 + 5t

this would mean normal accel. is (ds/dt)^2/R

I'm not sure what to evaluate in this?

is this done? do i have to go further?

@& x= 1 + 4t

y= 2 + 5t

is not a parameterization of a circle.

For a circle x^2 + y^2 is constant, and that is not the case for your x and y functions.

Any parameterization of the form x = +- r cos(f(t)), y = +_ r sin(f(t)) or x = +_ r sin(f(t)), y = +- r cos(f(t) would parameterize the circle. The simplest might be either

x = r cos(t), y = r sin(t), with period pi

or

x = r cos(2 pi t), y = r sin(2 pi t), with period 1.*@

@& Your parameterization, which is not a parameterization of the circle, would lead to r ' (t) = 4 `i + 5 `j and r '' ( t ) = 0. The acceleration would be zero, and there would be zero tangential and normal components. This is because your parameterization is a straight line, with no change in direction, and because the speed would always be sqrt( 4^2 + 5^2) = sqrt( 41 ).*@

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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:

R'(1) is OK which is V(t)

R''(1) is OK which is A(t)

N(t) is || R' X R'' || / || R' || which came out to be 0.6i + 0j - 0.6k / 45.45

@&

`r ' (t) = `v(t) = 3 cos(t) `i + 4 `j - 3 sin(t) `k

`r '' ( t) = `a(t) = -3 sin(t) `i - 3 cos(t) `k

*@

proj_V(1)(A(1)) is ( (v dot a) / v^2) * v which simplifies to -3( tan(t), 0, cot(t) ) at (t=1) = (-0.052, 0, 171.87)

proj_A(1)(N(1)) is ( ( (N dot A) ) / N^2 ) * N which simplifies to (-.05i, 0j, 2.96k)

sum of those is (-0.104i, 0j, 174.83k)

proj v(1) (a(1)) relates to A_t very close because A_T is V dot A / || V ||. the only difference in the projection is that the denominator is V^2 rather than magnitude

proj_N(1)(A(1)) relates to A_N. where A_N is V^2/R. which is (173.1i, 4j, -0.00083k). the relation is the i component? i'm having trouble figuring out the ralation between this as far as the numbers and direction.

I know A_N is normal acceleration which is the component back to the center of the circle in which your traveling around at that instant. if you project the normal component onto the acceleration. you should get a Binormal?

I'm still fuzzy on this one.

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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:

B represents Binormal vector. if a tangential vector is crossed with a Normal vector, then you get the binormal vector.

likewise, if you take the rate at which the normal vector changes, and cross it with the tangential vector (which doesn't change at that position) then you will get the rate at which the binormal vector changes.

I understand this fully, but as far as ""proving"" this, i'm having trouble starting this.

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