#$&* course Mth 277 qa 10_04The unit tangent vector is the unit vector in the direction of the velocity.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: `q002. What are the corresponding velocity and acceleration vectors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `R(t)=Acos(omega*t)`i+Asin(omega*t)`j 'V(t)=-A*omega*sin(omega*t)'i+A*omega*cos(omega*t)'j 'A(t)=-A*omega^2*cos(omega*t)'i-A*omega^2*sin(omega*t)'j confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: `q003. What is the speed of the point and what is the magnitude of the acceleration? Does either change with respect to t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `q005. Sketch the position vector, the velocity vector and the acceleration vector at the instant when omega * t = pi / 6. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: `q006. Let v be the speed of the point and r the radius of the circle. What is the expression for the magnitude of the acceleration in terms of v and r? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a_cent=v^2/r
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: `q007. If another point is moving around a different circle at a different constant speed v, then if the magnitude of the acceleration of that point is a_cent, what is the radius of the circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r = v^2 / a_cent confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe the point would once again have a circular path, but would be moving quicker. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: `q009. Find the acceleration and velocity vectors, and the magnitudes of both. Does either change with respect to t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `R(t) = A cos(omega * t^2) `i + A sin(omega * t^2) `j 'V(t)=-A2t*omega*sin(omega*t^2)i+A2t*omega*cos(omega*t^2)j 'A(t)=-A4t^2*omega*cos(omega*t^2)i+-A4t^2*omega*sin(omega*t^2)j Yes, both change with respect to the value of t.
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Given Solution: OK &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `q010. Find the components of the acceleration in the direction of the velocity, and perpendicular to the direction of the velocity. Find the magnitudes of both components of the acceleration. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am unsure how to solve this but I think you use projections.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `q011. If an object was moving around a circle with constant speed equal to that of this particle, with acceleration toward the center of the circle equal in magnitude to the perpendicular component of the acceleration, what would be the expression for the radius of that circle? Of course we know that the circle in this example has radius A, but don't use that knowledge in your solution. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r = v^2 / a_cent confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is a statement, not a question. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
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