QA10_4mth

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course Mth 277

qa 10_04The unit tangent vector is the unit vector in the direction of the velocity.

The tangential component of the acceleration is the projection of the acceleration vector on the unit tangent vector.

The normal component of the acceleration is the acceleration, minus its component in the direction of the unit tangent vector.

The unit normal vector is the unit vector in the direction of the normal component of the acceleration vector. A vector parallel to the normal vector can also be obtained by taking the derivative of the unit tangent vector.

The unit binormal vector is the cross product of the unit tangent and unit normal vectors.

The speed of a point whose position function is R(t) is the magnitude of the velocity vector.

Section 10.4

... A point moves around a circle of radius A at constant speed v. If the radial line vector from the center of the circle to the point

The position vector of a moving point is `R(t) = A cos(omega * t) `i + A sin(omega * t) `j. For this position function:

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Question: `q001. Sketch the path of the point.

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Your solution:

Well we can not sketch the exact path of the point because the radius is an unknown and we do not know the center of the circle. However, the path of the point cn be projected to be a circle which would have a radius of A.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Question:

`q002. What are the corresponding velocity and acceleration vectors?

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Your solution:

`R(t)=Acos(omega*t)`i+Asin(omega*t)`j

'V(t)=-A*omega*sin(omega*t)'i+A*omega*cos(omega*t)'j

'A(t)=-A*omega^2*cos(omega*t)'i-A*omega^2*sin(omega*t)'j

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Question:

`q003. What is the speed of the point and what is the magnitude of the acceleration? Does either change with respect to t?

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Your solution:

The speed of the point is sqrt((-A*omega*sin(omega*t))^2+(A*omega*cos(omega*t))^2)

The magnitude of the acceleration is sqrt((-A*omega^2*cos(omega*t))^2-(A*omega^2*sin(omega*t))^2)

Both of these equations would change with respect to t.

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Both of these expressions simplify. One of them simplifies to omega^2 * A and the other simplifies to something equally simple.

The results provide an important characterization of the speed and the acceleration involved in motion of this type.

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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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Question:

`q004. What is the angle between `R(t) and the velocity vector, and what is the angle between the velocity vector and the acceleration vector?

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Your solution:

`R(t)=Acos(omega*t)`i+Asin(omega*t)`j

'V(t)=-A*omega*sin(omega*t)'i+A*omega*cos(omega*t)'j

'A(t)=-A*omega^2*cos(omega*t)'i-A*omega^2*sin(omega*t)'j

To find the angle I do

cos(theta)=(R(t)dotV(t))/(magn(R(t))magn(V(t)))

theta between Rt Vt=arccos(-A*omega^2*cos(omega*t))(-A*omega*sin(omega*t))+(Asin(omega*t))(A*omega*cos(omega*t)))/((sqrt(Acos(omega*t))^2+(Asin(omega*t))^2)(sqrt((-A*omega*sin(omega*t))^2+(A*omega*cos(omega*t))^2)))

theta between Vt At=arccos(((-A*omega*sin(omega*t))(-A*omega^2*cos(omega*t))+(A*omega*cos(omega*t))(-A*omega^2*sin(omega*t))/((sqrt((-A*omega*sin(omega*t))^2+(A*omega*cos(omega*t))^2))(sqrt((-A*omega^2*cos(omega*t))^2-(A*omega^2*sin(omega*t))^2))))

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Good, but these expressions simplify considerably.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question:

`q005. Sketch the position vector, the velocity vector and the acceleration vector at the instant when omega * t = pi / 6.

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Your solution:

`R(t)=A*.866`i+A*.5`j

'V(t)=-A*omega*.5'i+A*omega*.866'j

'A(t)=-A*omega^2*.866'i-A*omega^2*.5'j

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Question:

`q006. Let v be the speed of the point and r the radius of the circle. What is the expression for the magnitude of the acceleration in terms of v and r?

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Your solution:

a_cent=v^2/r

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This is so, but how is this obtained?

It can be obtained from your earlier results, provided they are simplified.

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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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Question:

`q007. If another point is moving around a different circle at a different constant speed v, then if the magnitude of the acceleration of that point is a_cent, what is the radius of the circle?

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Your solution:

r = v^2 / a_cent

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Question:

The position vector of a moving point is `R(t) = A cos(omega * t^2) `i + A sin(omega * t^2) `j.

`q008. Sketch the path of the point.

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Your solution:

I believe the point would once again have a circular path, but would be moving quicker.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Question:

`q009. Find the acceleration and velocity vectors, and the magnitudes of both. Does either change with respect to t?

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Your solution:

`R(t) = A cos(omega * t^2) `i + A sin(omega * t^2) `j

'V(t)=-A2t*omega*sin(omega*t^2)i+A2t*omega*cos(omega*t^2)j

'A(t)=-A4t^2*omega*cos(omega*t^2)i+-A4t^2*omega*sin(omega*t^2)j

Yes, both change with respect to the value of t.

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Both expressions need to be simplified. The simplified expression for v is quite simple and revealing. The simplified expression for A is a little more complicated, but not too difficult to interpret.

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confidence rating #$&*:

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Given Solution: OK

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Self-critique (if necessary):

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Question:

`q010. Find the components of the acceleration in the direction of the velocity, and perpendicular to the direction of the velocity. Find the magnitudes of both components of the acceleration.

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Your solution:

I am unsure how to solve this but I think you use projections.

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That's correct.

The projection of one vector onto another is its component parallel to that vector.

If you subtract that projection from the vector, you get the perpendicular component.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question:

`q011. If an object was moving around a circle with constant speed equal to that of this particle, with acceleration toward the center of the circle equal in magnitude to the perpendicular component of the acceleration, what would be the expression for the radius of that circle? Of course we know that the circle in this example has radius A, but don't use that knowledge in your solution.

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Your solution:

r = v^2 / a_cent

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Question:

`q012. For any position function `R(t), we can follow the same procedure to find the radius of a hypothetical circle.

At any t, we can sketch the osculating circle, which is a circle whose radius is given by the expression found in the preceding question, subject to the condition that the direction of the radial vector (i.e., the vector from the center of that circle to the moving point) is in the direction opposite the unit normal. This is easily understood from a sketch. The center and radius of the osculating circle. The path of the particle at The curvature of the path at that instant is equal to the reciprocal of the radius of the osculating circle.

The osculating circle is the circle of the radius you found in the preceding, whose center

`q013. The curvature of the path of the particle is equal to the reciprocal of the radius of the osculating circle.

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Your solution:

This is a statement, not a question.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Question:"

Self-critique (if necessary):

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Self-critique rating:

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You really need to simplify a lot of those expressions.

It's fairly easy to do so. A lot of stuff factors out, and what's left is often equal to 1 by the Pythagorean Identity.

Revision is your option, but I do recommend that you go through those simplifications. If you can't get them, or aren't sure, you should submit at least some revisions.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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