What is the derivative of #f(t) = (tcos^2t , t^2sin^2t-cost ) #?

Answer 1

Emily Ammerman

#(dy)/(dx)=(t^2cost+2tcost+sint)/(cos^2t+2tsintcost)#

In parametric form of equation #f(t)=(x(t),y(t))#

#(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

Here #x(t)=tcos^2t# and #y=t^2sint-cost#

and therefore #(dx)/(dt)=cos^2t-txx2costxx(-sint)#

= #cos^2t+2tsintcost#

and #(dy)/(dt)=t^2cost+2tcost+sint#

Hence #(dy)/(dx)=(t^2cost+2tcost+sint)/(cos^2t+2tsintcost)#

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Answer 2

Ezekiel Alexander

The derivative of ( f(t) = (t\cos^2 t, t^2\sin^2 t - \cos t) ) with respect to ( t ) is:

[ f'(t) = \left( \frac{d}{dt}(t\cos^2 t), \frac{d}{dt}(t^2\sin^2 t - \cos t) \right) ]

Using the product rule and chain rule, the derivatives are:

[ f'(t) = \left( \cos^2 t - 2t\sin t \cos t, 2t\sin^2 t + 2t^2\sin t \cos t + \sin t \right) ]

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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.