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A rotating spotlight is shining on point P(x,y) on a shoreline and the portion of the shoreline on which the spotlight shines is in the shape of the parabola y=x^2 from the point (1,1) to the point (5, 25).

If the spotlight is rotating at the rate of one revolution per minute, how fast in the point P traveling along the shoreline at the instant it it at the point (3,9)???


Please help, anyone!!!! it is due tomorrow!!!!!


(from the free-response section of AP Calculus BC exam of 1996.)

By Bubbloy (Bubbloy) on Wednesday, April 14, 2004 - 03:47 am: Edit

so you set it up with the light coming from the origin, the tan function is appropriate for this question. the height y of the light on the island, x^2, is the leg of a triangle and the position of the dot on the island is x, the base of a triangle.
i will use @ for theta.

tan@=y/x
we know that d@/dt=2pi

at (3,9) tan@ = 3, so P=(tan@, tan@^2)

dP/dt= (sec@^2(d@/dt),(2sec@^2)(tan@)(d@/dt))

plug in d@/dt=2pi

avoiding some simple simplifications in the triangle.

dP/dt=(10*2pi,120pi)
so from physics, vector, w/e we get dP/dt=sqrt((dx/dt)^2+(dy/dt)^2)

so sqrt(400pi^2+14400pi^2)= ~382.1912416 ft/sec.

i feel your pain staying up until 4 am doing homework night break ends.

By Hankinator3000 (Hankinator3000) on Thursday, April 15, 2004 - 12:47 am: Edit

hahah thanks a lot, i felt stupid cuz i figured it out by myself within 10 minutes after i post this, but thank you very very much anyways!!! by the way, are you going to CPW?

By Hankinator3000 (Hankinator3000) on Thursday, April 15, 2004 - 12:51 am: Edit

interesting profile you have there too

By Bubbloy (Bubbloy) on Wednesday, April 21, 2004 - 01:17 am: Edit

i'm a junior, so no i didn't, been there though for a cpw type stay at phi sig, if all goes well, i will this time next year.