These are some resources for the sheet we worked on in class on Thursday.

Problem 1

(a)




problem a
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(b)

I begin by graphing the distance formula with the parametric equations substituted.



I then look for the MINIMUM on the graph. The coordinates will be (time,minimum distance).

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(c)

You can answer this from the applet above.

(d)

problem d
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(e)

problem e
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Problem 2


prob2triangle.png
Sketch in the right triangles as shown on the graph at left. You can repeatedly use INVERSE TAN to find the marked angles. Once you have the marked angles, it is simple angle chasing to find the desired angles (either subtract the angles from 90 or subtract the angles from 180).

Try and generalize what is happening. These right triangle are nothing more than SLOPE TRIANGLES, so the angles in the triangle are related in some way to the slopes of the sides of the triangles.



Problem 3


(a)

(b)

The length of the vector is 13 units. Therefore, a vector that is 1-unit long and pointing in the opposite direction can be found by multiplying by the appropriate scalar.



(c)

There are two possible answers for this, depending on which perpendicular vector you choose (and you should choose both!). Multiplying by the scaler 3, and flipping the components and changing the signs of the correct component, we have [36,15] or [-36,-15].

(d)

If my math is correct, A'(8,-15), B'(4,-10), and C'(12,-7)

(e)

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Problem 4


(a)

(0.41176,1.64706)

(b)

(5,0.5)

(c)

(1.6666666,1.3333333)

(d)

(11/3, 8/3)

(e)

(11/3, 8/3)