Christmas Streak 53/88: Calculus 1 Basics #3

Calculus Level 2

A point P ( x , y ) P(x,~y) moves around in the coordinates plane with the parametric relation

{ x = t 4 + 1 t y = t 2 + 2 t . \large \cases{x = t^4+\frac{1}{t} \\\\ y=t^2+2t}.

Find the speed at which P P moves at t = 1. t=1.

This problem is a part of <Christmas Streak 2017> series .


The answer is 5.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Nov 23, 2017

d x d t = 4 t 3 − 1 t 2 d y d t = 2 t + 2 \frac{dx}{dt}=4t^3-\frac{1}{t^2} \\\\ \frac{dy}{dt}=2t+2

So the velocity of P P at t = 1 t=1 is ( 3 , 4 ) . (3,~4).

Therefore, the speed of P P at t = 1 t=1 is 3 2 + 4 2 = 5 . \sqrt{3^2+4^2}=\boxed{5}.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You got a typo in your first equation, it's actually d x d t = 4 t 3 − t − 2 \frac{dx}{dt}=4t^3-t^{-2} or d x d t = 4 t 3 − 1 t 2 \frac{dx}{dt}=4t^3-\frac{1}{t^2} but your typo leads to the same answer

Alexander Ilmexia - 3 years, 4 months ago

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Indeed x'D

Thank you!

Boi (보이) - 3 years, 4 months ago

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