How do you differentiate the following parametric equation: $x(t)=lnt/t, y(t)=(t-3)^2$ ?

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Answer 1 · 2017-03-04T10:51:54

$dy/dx={2t^2(t-3)}/(1-lnt), t!=0,t!=e.$

By the Rule of Parametric Diffn., $dy/dx=(y'(t))/(x'(t)), x'(t)!=0.$

Now, using the Chain Rule,

$y'(t)=d/dt{(t-3)^2}=2(t-3)d/dt{(t-3)}.$

$:. y'(t)=2(t-3)........(1).$

Next, $x(t)=lnt/t$

$:. x'(t)={td/dt(lnt)-(lnt)d/dt(t)]/t^2...."[The Quotient Rule]"$

$={t(1/t)-(lnt)(1)]/t^2$

$:. x'(t)=(1-lnt)/t^2..........(2).$

Also, let us note that,

$x'(t)=0 rArr t=e, or, t!=e rArr x'(t)!=0.........(2').$

Altogether, taking into account $(1),(2), and (2'),$

$dy/dx={2(t-3)}/{(1-lnt)/t^2}$

$:. dy/dx={2t^2(t-3)}/(1-lnt), t!=0,t!=e.$

Enjoy Maths.!

How do you differentiate the following parametric equation: x(t)=lnt/t, y(t)=(t-3)^2 x(t)=lnt/t, y(t)=(t-3)^2 ?