Calc AB Question





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Discus: SAT/ACT Tests and Test Preparation: May 2003 Archive: Calc AB Question
By Parhelia (Parhelia) on Thursday, May 01, 2003 - 09:44 pm: Edit

Go here:
http://www.collegeboard.com/ap/pdf/calculus_ab_01.pdf
to see the 2001 Calc AB free response section. Go to question (1) and do part (c)

I want to see if my answer is correct (7.292 units^3).

By Heatwave345 (Heatwave345) on Thursday, May 01, 2003 - 09:58 pm: Edit

ummm... i got 8.332

i used the "disk" method and ended up with:

pi*integral(((2-x^3)^2-(tan(x))^2)dx) from 0 to 0.902

0.902 is the intersection of the two functions.

plugging that into the calculator yields 8.332 units^3

By Parhelia (Parhelia) on Thursday, May 01, 2003 - 10:22 pm: Edit

Heatwave, you're correct, the answer is 8.332 units^3. I see what I did wrong.

By Parhelia (Parhelia) on Sunday, May 04, 2003 - 07:45 pm: Edit

I have another calc question.

Go here:
http://collegeboard.com/prod_downloads/ap/students/calculus/calculus_ab_frq_02.pdf
and do number (5). These are my (probably wrong) answers:

a) 125pi/12 cm^3
b) -15pi/8 cm^3/hr
c) could someone explain this? Is the constant of proportionality equal to -3/10?

By Brd (Brd) on Sunday, May 04, 2003 - 08:26 pm: Edit

All your answers are correct. To show part c), you just have to show that

A = pi r^2 = (1/4) pi h^2

and

dV/dt = -(3/40) pi h^2

so that

dV/dt = -(3/10) [(1/4) pi h^2] = -3/10 A

and you are done.

By Parhelia (Parhelia) on Sunday, May 04, 2003 - 08:36 pm: Edit

thanks!


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