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.!