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AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

Sunday, November 27, 2005

Friday's class...

Last Friday we started our class by having our second review quiz in preparing for the Test in May. We're given 8 minutes to answer 4 multiple type questions. Then after that Mr. K showed us the rubric for the Prince of Calculand Project. He discuss it to us for a few minutes. After that Mr. K talked about the Editor's Initiative and he asks our thoughts about it.

After all the discussions we had about the project and the Editor's Initiative, Mr. K showed us and taught us how to answer this related problem that we have been working for a few days.

A spotlight at a school dance is fastened to a wall 8 m above the floor. A girl 1.75 m tall moves away from the wall at a speed of 0.75 m/s.
a. At what rate is the length of her shadow increasing?
b. At what speed is the tip of her shadow moving?

First in solving in this problem like other kind of problems is drawing a diagram on what is given to you.

Then next is analyze the drawing. In our diagram there are two similar triangles and based on that we could create a relationship between the two.
b/1.75= a+b/8 (a+b=c as shown in the diagram)

8b=1.75a+1.75b
6.25b=1.75a

Then find the derivative of that,
6.25 db/dt= 1.75da/dt
Since we know da/dt= 0.75 m/s, juss plug it in and we could figure out db/dt and justplug it in.

db/dt=1.75(0.75)/6.25
=0.21 m/s
So the length of the girl's shadow is increasing 0.21 m/s.

Then in order to know what speed is the tip of her shadow. We could use this relationship c=a+b and find the derivative of that,
dc/dt=da/dt+db/dt and plug in the values to get dc/dt
dc/dt=(0.75)+(0.21)
dc/dt=
0.96 m/s

The tip of the girl's shadow is moving 0.96 m/s.

That's all we did last Friday....The next scribe is Ara..I'm willing to accept any constructive criticism so don't be afraid to leave a comment. Thanks!




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