#$&* course Mth 277 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `v(t) = `R ' (t) = -cos(t) `i + sin(t) `j + `k `a(t) = `v ' (t) = `r '' (t) = sin(t) `i + cos(t) `j &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Is there a reason why you did the integral instead of the derivative for the velocity? Or did I mess something up?
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Given Solution: Divide `v(t) by || `v(t) || and simplify &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I tried to simplify and I could not seem to simplify it. I believe the reason why is because our velocity vectors are different. ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q003. What is the component of the acceleration vector in the direction of the unit tangent vector? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The component of the acceleration vector in the direction of the unit tangent vector is the projection of the acceleration vector onto the unit tangent vector. (-sin(t)i-cos(t)j dot (cos(t)-sin(t)+1)/(sqrt(cos^2(t)-sin^2(t)+1^2)))/(-sin(t)i-cos(t)j)
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Given Solution: The component is denoted `a_T (t) . The desired component is the projection of `a(t) on `T(t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q004. What is the component of the acceleration vector in the direction perpendicular to the unit tangent vector? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find this, we need to do `a(t) - `a_T(t). The result of that gives us the vector a_N(t).
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Given Solution: Subtract the component `a_T(t) from `a(t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q005. What is the normal component of the acceleration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The unit normal vector is the unit vector in the direction of `a_N(t), and is perpendicular to the unit tangent vector. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This is the component perpendicular to the unit tangent vector. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q006. Show that the normal component of the acceleration is perpendicular to the tangential component. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To do this, we would take the normal componenet of the acceleration and dot it by the tangential component. If there result is 0 then the vectors are perpendicular. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Two vectors are perpendicular if their dot product is zero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q007. Show that the direction of the derivative of the unit tangent vector is the same as that of the unit normal vector. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The direction of the derivative of the unit tangent vector is the same as the the unit normal vector if cos(theta)=0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Two vectors are parallel if the cosine of the angle between them is zero. How therefore can to test to see if the vectors are parallel? What further test allows us to determine if they are in the same direction, vs. in the opposite directions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. Find the unit normal vector. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The unit normal vector is the derivative of the unit tangent vector divided by the derivative of its magnitude. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You have at least one vector in the normal direction (in fact in the preceding questions you have found two). Use either to find the unit normal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK *********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q010. What difference would it make in the above results if the function was `R(t) = sin(t^2) `i + cos(t^2) `j + t^2 `k? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would change the values quite a bit. This is because the velocity and acelleration vectors would change drastically since the chain rule would be used for the derivative and a domino effect would cause the other values to change as well. However, the same concepts will be used as established above. Also, there would be a k component to the velocity vector of this problem since the k component of the position vector is t^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!