Query10_5mth

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course Mth 277

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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:

R(t)=<3/5cos(t),4/5(1+sin(t),cost>

V(t)=<-3/5sin(t),45cos(t),-sin(t)>

a(t)=<-3/5cos(t),-4/5sin(t),-cos(t)>

||v(t)||=sqrt(9/25sin^2(t)+16/25cos^2(t)+sin^2(t)

=sqrt(16/25+18/25sin^2(t))

a_t=<-3/5sin(t),45cos(t),-sin(t)> dot <-3/5cos(t),-4/5sin(t),- cos(t)>/sqrt(9/25sin^2(t)+16/25cos^2(t)+sin^2(t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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 =VdotA/sqrt(V)

a_t=(-37) / sqrt(25 + 4 + 16)

a_t=-37 / sqrt(45)

a_t= -37sqrt(5) / 15

a_t=-5.5

a_n=||(v X a)|| / ||v||

a_n=||<5,-2,4> X <1,3,-9>|| / sqrt(45)

a_n=||<6, -49, 17>|| / 3sqrt(5)

a_n=sqrt(2726) / 3sqrt(5)

a_n=7.8

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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:

x^2 + y^2 = r^2

x=rcos(theta)

y=rsin(theta)

r(t)=rcos(theta)i + rsin(theta)j

v(t)=-rsin(theta)i + rcos(theta)j

a(t)=-rcos(theta)i - rsin(theta)j

||v(t)|| = sqrt[r^2sin(theta) + r^2cos(theta)]

v(t) dot a(t) / ||v(t)|| = <-rsin(theta)i + rcos(theta)j> dot <-rcos(theta)i - rsin(theta)j> / r

= r^2sin(theta)cos(theta) -r^2sin(theta)cos(theta) / r

@&

Your numerator should be grouped and the expression simplified, which will give you 0.

a(t) has 0 component in the direction of v(t)

This is a very important property of circular motion at constant speed.

*@

a_n = ||v X a|| / ||v||

= ||r^2sin(theta) + r^2cos(theta)|| / r

= sqrt(r^4) / r

= r

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:

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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:

V(t)=<3cos(t),4,-3sin(t)>

V(1)=<1.62,4,-2.52>

A(t)=<-3sin(t),0,-3cos(t)>

A(1)=R''(1)=<-2.52,0,-1.62>

T(t)=<3cos(t),4,-3sin(t)>/(sqrt((3cos^2(t)+16-3sin^2(t))))

N(t)=T'(t)/||T'(t)||

A(1) onto V(1)

=(V(1) dot A(1))V(1)/(||V(1)||^2)

proj_V(1) (A(1))=<0,0,0>

I believe when you add together these two projections you get R""(1).

Since proj_V(1) (A(1))=<0,0,0>

proj_N(t)(A(1))=<-2.52,0,-1.62>

@&

V(t) is the velocity function, which is the time derivative of the position function:

V(t) = R ' (t).

T(t) is the unit tangent vector. The velocity V(t) is tangent to the curve defined by position vector R(t), so if we divide V(t) by its magnitude we get a unit vector tangent to the curve. So

T(t) = V(t) / || V(t) || = R ' (t) / || R ' (t) ||.

N(t) is the unit normal vector, perpendicular to the tangent. The derivative of a unit vector is perpendicular to that vector. T(t) is a unit vector tangent to the curve, so T ' (t) is perpendicular to the tangent, and is therefore normal to the curve. Thus

N(t) = T ' (t) / || T ' (t) ||

is a unit vector normal to the curve.

A(t) = V ' (t) = R '' ( t ) is the acceleration vector, the second derivative of the position vector.

A(t) lies in the plane of T(t) and N(t), and components parallel to each of these vectors. ( A(t) is therefore a linear combination of T(t) and N(t) ).

The component of a given vector in the direction of another vector is its projection onto that vector

(proj_`w(`v) is the projection of `v onto `w, and has magnitude v cos(theta), where theta is the angle between `v and `w; thus proj_`w(`v) = v cos(theta) * `w / || w ||, where `w / || `w || is the unit vector in the direction of `w; since v cos(theta) = `v dot `w / || `w ||, the projection is `v dot `w * `w / || `w ||^2. If `w is a unit vector then || `w || = 1 and the projection is `v dot `w * `w).

Thus the component of A(t) in the direction of the unit tangent is A_T = A(t) dot T(t), while the component of A(t) in the direction of the unit normal is A_N = A(t) dot N(t). A(t) = A_T * T(t) + A_N * N(t) (a linear combination of the unit tangent and unit normal vectors with constant A_T and A_N).

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@& The above summary confuses, to an extent, vector and scalar projections. The component of one vector in the direction of another can be specified by the scalar projection

proj_`w (`v) = `v dot `w / || `w ||

or the vector projection

`proj_`w (`v) = `v dot `w / || `w || * `w / || `w ||.

The former is just the number that tells you what to multiply by the unit vector in the direction of `w to get the vector component.

The sum

`proj_`T ( `A) + `proj_`N ( `A)

of the vector components in the directions of the unit tangent and unit normal vectors (which we remember are perpendicular) is equal to the vector `A.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:

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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:

T and N perp unit vect

Therefore, there cross product is perp to both vectors

a_t=v(t) dot a(t)/||v(t)||

a_n=||v X a||/||v||

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

This one gets extremely confusing.

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Self-critique rating:"

&#Good responses. See my notes and let me know if you have questions. &#