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How do you differentiate the following parametric equation: # x(t)=1-lnt, y(t)= cost-t^2 #?

Answer 1

Isaiah Allen

#dx/dt = -1/t# and #dy/dt = -sin t -2t#

#dx/dt = -1/t# and #dy/dt = -sin t -2t#. If required to have #dy/dx# , it

would be #dy/dt/dx/dt#

= t(sin t +2t)

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Answer 2

Emma Aldridge

To differentiate the parametric equations x(t) and y(t), you would use the chain rule for differentiation. The chain rule states that if u = f(g(t)), then du/dt = f'(g(t)) * g'(t). Applying this rule to each equation separately:

For x(t) = 1 - ln(t), differentiate x(t) with respect to t: dx/dt = d(1 - ln(t))/dt = 0 - (1/t) = -1/t.

For y(t) = cos(t) - t^2, differentiate y(t) with respect to t: dy/dt = d(cos(t) - t^2)/dt = -sin(t) - 2t.

So, the differentiation of the parametric equations x(t) and y(t) are: dx/dt = -1/t dy/dt = -sin(t) - 2t.

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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.