Asked by Amber
A flower pot is dropped from a point 12 feet above the ground,so that it will land 4 feet from the base of a 12 foot lamppost. At what rate is the shodow of the flower pot moving along the ground when the pot is 8 feet above the ground?
Answers
Answered by
Steve
height y = 12 - 16t^2
shadow at x, similar triangles, so
x/12 = (x-4)/y
or,
xy = 12x - 48
y dx/dt + x dy/dt = 12 dx/dt
y=8 when t=1/2
x=6 when y=8
(12 - 16t^2) dx/dt + x (-32t) = 12 dx/dt
at t=1/2,
8 dx/dt - 96 = 12 dx/dt
dx/dt = 96/-4 = -24 ft/s
shadow at x, similar triangles, so
x/12 = (x-4)/y
or,
xy = 12x - 48
y dx/dt + x dy/dt = 12 dx/dt
y=8 when t=1/2
x=6 when y=8
(12 - 16t^2) dx/dt + x (-32t) = 12 dx/dt
at t=1/2,
8 dx/dt - 96 = 12 dx/dt
dx/dt = 96/-4 = -24 ft/s
Answered by
Steve
Oops.
x=12 when y=8
make the change and percolate it through the rest of the solution.
x=12 when y=8
make the change and percolate it through the rest of the solution.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.