How do you find #dy/dx# for the curve #x=t*sin(t)#, #y=t^2+2# ?

Answer 1

Emilia Aguilar

To find the derivative of a parametric function, you use the formula:

#dy/dx = (dy/dt)/(dx/dt)#, which is a rearranged form of the chain rule.

To use this, we must first derive #y# and #x# separately, then place the result of #dy/dt #over #dx/dt#.

#y=t^2 + 2#

#dy/dt = 2t# (Power Rule)

#x=tsin(t)#

#dx/dt = sin(t) + tcos(t)# (Product Rule)

Placing these into our formula for the derivative of parametric equations, we have:

#dy/dx = (dy/dt)/(dx/dt) = (2t)/(sin(t)+tcos(t))#

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Answer 2

Charles Arocha

To find ( \frac{dy}{dx} ) for the curve ( x = t \sin(t) ) and ( y = t^2 + 2 ), differentiate ( y ) with respect to ( x ) using the chain rule:

[ \frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} ]

[ \frac{dy}{dt} = 2t ] [ \frac{dx}{dt} = \sin(t) + t\cos(t) ]

[ \frac{dy}{dx} = \frac{2t}{\sin(t) + t\cos(t)} ]

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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.