## - Parametric Differentiation

Given that $$x=f(t)$$ and $$y=g(t)$$ where $$t$$ is called a parameter, then

$$\displaystyle {dy\over dx}={dy\over dt}×{dt\over dx}$$

LESSON 1
Find the gradient of the stated curve at the point defined.
$$x=t^2 (4t-1); y=\sqrt{2t+7}$$ when $$t=1$$
SOLUTION
• Determine $$\displaystyle {dx\over dt}$$
$$x=t^2 (4t-1)$$
$$x=4t^3-t^2$$
$$\displaystyle {dx\over dt}=12t^2-2t$$
• Determine $${dy\over dt}$$
$$\displaystyle y=\sqrt{2t+7}$$
$$\displaystyle y=(2t+7)^{{1\over2}}$$
$$\displaystyle {dy\over dt}={1\over2} (2t+7)^{-{1\over2}}(2)$$
$$\displaystyle =(2t+7)^{-{1\over2}}$$
• Determine $$\displaystyle {dy\over dx}$$ as $$\displaystyle {dy\over dt}×{dt\over dx}$$
$$\displaystyle {dy\over dx}={dy\over dt}÷{dx\over dt}$$
$$\displaystyle ={(2t+7)^{-{1\over2}}\over 12t^2-2t}$$
$$\displaystyle ={1\over 2(6t^2-t) \sqrt{2t+7}}$$
• Substitute the given value of t into the formula for the gradient of the curve.
When $$t=1$$
$$\displaystyle {dy\over dx}={1\over 2(6(1)^2-(1))\sqrt{2(1)+7}}$$
$$\displaystyle ={1\over30}$$

LESSON 2
Find the equation of the normal to the curve $$\displaystyle x={8\over t^3}, y=2t^2-1$$ at the point where the curve crosses the line $$x=1$$.
SOLUTION
• Determine $$\displaystyle {dx\over dt}$$
$$\displaystyle x={8\over t^3}=8t^{-3}$$
$$\displaystyle {dx\over dt}=-24t^{-4}=-{24\over t^4}$$
• Determine $$\displaystyle {dy\over dt}$$
$$y=2t^2-1$$
$$\displaystyle {dy\over dt}=4t$$
• Determine $$\displaystyle {dy\over dx}$$, the formula for the gradient of the tangent to the curve.
$$\displaystyle {dy\over dx}={dy\over dt}×{dt\over dx}$$
$$=4t×\left(-{t^4\over 24}\right)$$
$$\displaystyle =-{t^5\over 6}$$
• Determine the value of $$t$$ when $$x=1$$
When $$x=1$$
$$\displaystyle x={8\over t^3}$$
$$\displaystyle 1={8\over t^3}$$
$$t^3=8$$
$$t=2$$
• Determine the corresponding value of $$y$$. This gives the coordinate of the point at which $$x=1$$.
When $$t=2$$
$$y=2t^2-1$$
$$y=2(2)^2-1$$
$$y=7$$
$$(1, 7)$$
• Determine the gradient of the tangent at the point $$(1,7)$$.
At (1, 7):
$$\displaystyle {dy\over dx}=-{t^5\over 6}$$
$$\displaystyle =-{2^5\over 6}$$
$$\displaystyle =-{16\over3}$$
• Determine the gradient of the normal to the curve at the point $$(1, 7)$$.
Gradient of normal is $$\displaystyle {3\over16}$$
• Determine the equation of the normal
$$y=mx+c$$
$$\displaystyle 7={3\over16} (1)+c$$
$$\displaystyle {109\over16}=c$$
$$\displaystyle y={3\over16} x+{109\over16}$$

LESSON 3
Find $$\displaystyle {dy\over dx}$$ and $$\displaystyle {d^2y\over dx^2}$$ for the parametric equations $$x=3t^2-1$$ and $$y=t^2-6t-3$$.
Hence find and classify the stationary point(s).
SOLUTION
• Determine $$\displaystyle {dx\over dt}$$
$$x=3t^2-1$$
$$\displaystyle {dx\over dt}=6t$$
• Determine $$\displaystyle {dy\over dt}$$
$$y=t^2-6t-3$$
$$\displaystyle {dy\over dt}=2t-6$$
• Determine $$\displaystyle {dy\over dx}$$
$$\displaystyle {dy\over dx}={dy\over dt}×{dt\over dx}$$
$$\displaystyle ={2t-6\over 6t}$$
$$\displaystyle ={t-3\over 3t}$$
• Implicitly differentiate $$\displaystyle {dy\over dx}$$ which is written in terms of $$t$$ with respect to $$x$$.
$$\displaystyle {d^2y\over dx^2}={1(3t)-(t-3)(3))\over (3t)^2}{dt\over dx}$$
$$\displaystyle ={3t-3t+9\over 9t^2}×{1\over 6t}$$
$$\displaystyle ={1\over 6t^3}$$
Stationary points occur when $$\displaystyle {dy\over dx}=0$$
$$\displaystyle {t-3\over 3t}=0$$
$$t-3=0$$
$$t=3$$
The stationary point occurs when $$t=3$$. With this value we can determine the corresponding $$x$$ and $$y$$ values of the stationary point.
$$x=3(3)^2-1=26$$
$$y=3^2-6(3)-3=-12$$
$$(26, -12)$$
Given that $$t=3$$, we can substitute this value into the expression for $$\displaystyle {d^2 y\over dx^2}$$ in order to determine the nature of the stationary point.
$$\displaystyle {d^2 y\over dx^2}={1\over 6(3)^3}={1\over 162}$$
Minimum point