Consider the parametric curve given by: $x=3\cos(t)$, $y=t^{3/2}$. The question asks to find the equation of the tangent to this curve at the point $t=\pi/4$.
I've determined $$\frac{dy}{dx} =(\frac{dy}{dt})/(\frac{dx}{dt}) = -0.222$$ Have I got the right idea?
Also asks for the solution to be in the form $y=mx+c$, thank you.
 A: first find the $x$ and $y$ coordinates of the point at time $t = \pi/4.$ we have $x(\pi/4) = 3\sqrt 2/2 = 2.121, y = \left(\frac{\pi}4\right)^{3/2} = 0.696.$
now evaluate the slope at this point  $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}\big|_{t = \pi/4} = \frac{-3\sin \pi/4}{3\sqrt 2/2} = -1$$ therefore the tangent line at $t = \pi/4$ is $$y-0.696 = -1(x -2.121)\to y = 2.817-x $$
A: The direction vector of the tangent you're looking for is given by $(\frac{\mathbb d x}{\mathbb d t},\frac{\mathbb d y}{\mathbb d t})$, evaluated in $\frac \pi 4$. Then you can determine $m$. Finally you deduce $c$ by taking $t = \frac \pi 4$ : since $(x(\frac \pi 4),y(\frac \pi 4))$ is a point of the tangent, $y(\frac \pi 4) = m x(\frac \pi 4) + c$.
A: Hints: The usual way of finding a tangent line to a parametric curve $(x(t), y(t))$ at some number $t_{0}$ is to compute


*

*A point on the line, such as the point of tangency $(x_{0}, y_{0}) = (x(t_{0}), y(t_{0}))$.

*The slope of the tangent line, which is $m = y'(t_{0})/x'(t_{0})$ if $x'(t_{0}) \neq 0$.
Then use the point-slope form
$$
y - y_{0} = m(x - x_{0}),\quad\text{or}\quad
y = mx + (y_{0} - mx_{0}).
$$
