Question

A kite 100 ft above the ground moves horizontally at a speed of 14 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?
is the answer 7/50...please someone help me..it is urgent!

Answer 1

How did you get the rate of change of the angle to be 7/50? Can you show your work?

Answer 2

sure

Answer 3

using pythagoras theory, I found x to be SQRT 30000....then I found theta...to be 0.5236 (in radians)...cos(theta) = x/200 and the derivative is -sin(theta)*dtheta/dt = 1/200 * dx/dt...plug all the values in and you'll get 7/50...note that dx/dt is -14 because it is reducing

Answer 4

What is your cos(theta)?

Answer 5

cos theta is x/200
= SQRT 30000/200

Answer 6

My bad, I mean your sin(theta) in your derivative equation.

Answer 7

0.5

Answer 8

Whoops my bad, I forgot to flip my 1/2. Yep, that looks right!

Answer 9

Yeah?...Igot it wrong in my webassign though..I don't know why!

Answer 10

I got*

Answer 11

Is it because you forgot to specify units?

Answer 12

Or is it because you forgot your negative sign (from when you took the derivative of cos -> -sin)? The rate of change should be negative (the angle should be getting smaller as you release more of your kite).

Answer 13

I tried negative too but it was wrong...and the units were specified too

Answer 14

That's weird. I have no idea why it wouldn't be -7/50 deg/sec though. Sorry about that.

Answer 15

yeah..I know right..thanks a bunch anyway!

Answer 16

Just for ref:
If A is the angle in question then: tan(pi/2 - A) = x/100
Take the derivative: -sec^2(pi/2 - A) dA/dt = (dx/dt)/100 = 14/100

dA/dt is what you want so:
dA/dt = -(14/100)/sec^2(pi/2 - A)
dA/dt = -(14/100)cos^2(pi/2 - A)

The cosine of pi/2 - A when the string is 200 ft is 100/200 = 1/2

dA/dt = -(14/100)(1/2)^2 = -7/200 @REPS

Answer 17

Ohhh...aiit! Thanks! but isn't the derivative of tan, sec^2 and not -sec^2

Answer 18

@koikkara