Question

HELP DIFFERENTIAL CALCULUS:
PLEASE SEE ATTACHED FILE

Answer 1

\(tan\theta = \dfrac{y}{x}\)
Take derivative both sides with respect to \(theta\) for RHS, w.r.t. t for LHS
You have
\(sec^2(\theta) d(\theta) = \dfrac{x(dy/dt) -y (dx/dt))}{x^2}dt\)

\(\dfrac{d(\theta}{dt}=\dfrac{x(dy/dt)-y(dx/dt)}{sec^2(\theta)x^2}\)

Now, calculate the denominator: \(sec^2(\theta) = 1+tan^2(\theta)= 1+(y^2/x^2)\)
hence denominator is \(x^2+y^2\) as what we want.
Done.

Answer 2

THANKS!!