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

Answer 1

Christopher Agresta

#dy/dx=t#

The derivative #dy/dx#of parametric equations is.

#color(red)(|bar(ul(color(white)(a/a)color(black)(dy/dx=(dy/dt)/(dx/dt))color(white)(a/a)|)))#

here x = lnt #rArrdx/dt=1/t#

and #y=t-3rArrdy/dt=1#

#rArrdy/dx=1/(1/t)=1xxt/1=t#

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Answer 2

Charles Arocha

To differentiate the parametric equations (x(t) = \ln(t)) and (y(t) = t - 3), you differentiate each equation separately with respect to (t) using the chain rule for (x(t)) since it involves a natural logarithm:

[ \frac{dx}{dt} = \frac{1}{t} ]

and for (y(t)):

[ \frac{dy}{dt} = 1 ]

So, the parametric equations differentiate to:

[ \frac{dx}{dt} = \frac{1}{t} \quad \text{and} \quad \frac{dy}{dt} = 1 ]

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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.