## yamyam70 one year ago HELP DIFFERENTIAL CALCULUS: PLEASE SEE ATTACHED FILE

1. yamyam70

2. anonymous

$$tan\theta = \dfrac{y}{x}$$ 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.

3. yamyam70

THANKS!!