tan(q) = y/x
sec^2(q)dq/dt = [xdy/dt - ydx/dt]/x^2
when x = 2 = y q = pi/4 seq(q) = sqrt(2)
2dq/dt = [-1/2-2]/4
dq/dt = -5/16 rad/s
Hello, can someone please help me to set this problem up. Let theta(in radians) be an acute angle in a rt triangle, and let x and y respectively be the lengths of the sides adj and opp to theta. Suppose x and y vary with time.
At a certain instant, x =2 units and is increasing at 1 unit/s, while y = 2 units and is decreasing at 1/4 unit/s. How fast is theta chaning at that instant?
I know the answer is d0/dt = cos^0/x^O(x(dy/dt) - y(dx/dt) where I am representing 0 as theta.
Thanks,
Pantera