Given #y=(x/5)^2# how do you derive a parametric equation?

Answer 1

Christopher Agresta

The simplest way is to let #x = t#, then #y = (t/5)^2# and the parametric point is #(t,(t/5)^2)#.

Please see the graphs in the explanation.

The following is a graph of the equation #y = (x/5)^2#

The following is a graph of the parametric point #(t,(t/5)^2)#

Please observe that the graphs are identical.

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Answer 2

Emilia Aguilar

To derive parametric equations from the given equation ( y = \left(\frac{x}{5}\right)^2 ), we can introduce a parameter ( t ) and express ( x ) and ( y ) in terms of ( t ). Let ( x = 5t ). Substituting this into the original equation yields:

[ y = \left(\frac{5t}{5}\right)^2 = t^2 ]

So, the parametric equations are ( x = 5t ) and ( y = t^2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.