Implicit differentiation & parametricequations

Hello, Ley!

Quote:

Assume that the equations in each part define x and y implicitly as differentiable functions, $x=f(t),\:y=g(t)$ .
Find the slope of the curve $x=f(t),\:y=g(t)$ at the given value of $t.$

$a)\:\begin{Bmatrix}x^2-2tx+2t^2&=&4 & [1] \\2y^3-3t^2&=&4 & [2] \end{Bmatrix}\quad t=2$

Differentiate implicitly.

$[1]\;\;2x\frac{dx}{dt} - 2t\frac{dx}{dt} - 2x + 4t \:=\:0 \quad\Rightarrow\quad 2x\frac{dx}{dt} - 2t\frac{dx}{dt} \:=\:2x - 4t$

. . . $2(x-t)\frac{dx}{dt} \;=\;2(x-2t) \quad\Rightarrow\quad \frac{dx}{dt} \:=\:\frac{x-2t}{x-t}$

$[2]\;\;6y^2\frac{dy}{dt} = 6t \:=\:0 \quad\Rightarrow\quad \frac{dy}{dt} \:=\:\frac{t}{y^2}$

$\frac{dy}{dx} \;=\;\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \;=\;\frac{\dfrac{t}{y^2}}{\dfrac{x-2t}{x-t}} \;=\;\frac{t(x-t)}{y^2(x-2t)}$ .[3]

Substitute $t = 2$ into [1]:
. . $x^2 - 4x + 8 \:=\:4 \quad\Rightarrow\quad x^2 - 4x + 4 \:=\:0 \quad\Rightarrow\quad (x-2)^2 \:=\:0 \quad\Rightarrow\quad x \:=\:2$

Substitute $t = 2$ into [2]:
. . $2y^3 - 12 \:=\:4 \quad\Rightarrow\quad 2y^3 \:=\:16 \quad\Rightarrow\quad y^3 \:=\:8 \quad\Rightarrow\quad y \:=\:2$

We have: . $\begin{Bmatrix}t = 2 \\ x = 2\\ y = 2\end{Bmatrix}$

Substitute into [3]: . $\frac{dy}{dx} \;=\;\frac{2(2-2)}{2^2(2-4)} \;=\;0$

At the point $(2,2)$ , the slope is $0.$