|By Hankinator3000 (Hankinator3000) on Wednesday, April 14, 2004 - 02:58 am: Edit|
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.
we know that d@/dt=2pi
at (3,9) tan@ = 3, so P=(tan@, tan@^2)
plug in d@/dt=2pi
avoiding some simple simplifications in the triangle.
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.
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