Question

Calculus Related Rates HELP!

Answer 1

On Halloween night, Charlie Brown has been trying to catch the Great Pumpkin as it
rises from the pumpkin patch to bring all good calculus students presents and candy,
so he can keep all the goodies for himself. Charlie spots the Great Pumpkin just
starting to rise at a distance of 80 feet away. If the angle of elevation of the great
Pumpkin is changing at the rate of 0.5 radians per second,

Answer 2

a) How fast is the distance off the ground of the great Pumpkin changing when
the great Pumpkin is 6 feet off the ground?

Answer 3

I'm looking for dy/dt, right like the rate of change of the altitude/height

Answer 4

so we have the equation

tan (theta ) = y/80

80 tan (theta ) = y

take derivative with respect to time, both sides

80 * sec^2(theta) * dtheta/dt = dy/dt

Answer 5

dtheta /dt was given , 0.5 rad/sec

and you need to find theta,
you can use tan(theta) = 6/80 , and solve for theta

Answer 6

wait, why is the sec squared?