- 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