Parametric equation gives dy/dt = sin(t) +3 and x(t)= 6t^2+ln(t) and asks for when the graph of the position will have a horizontal tangent

Question

To my understanding if $\dfrac{dy}{dt}=0$ then the tangent is horizontal, but $\sin(t)+3$ will never equal $0$, therefore there is no horizontal tangent. Thanks!

Answer 1

You cannot conclude so fast $$x=6 t^2+\log (t) \qquad \qquad \frac{dx}{dt}=12 t+\frac{1}{t}$$ $$\frac{dy}{dx}=\frac{\sin (t)+3}{12 t+\frac{1}{t}}$$ What happens when $t \to 0$ ?