# How do you find the equation of the tangent line to the curve given by parametric equations: #x=1+(1/t^2)#, # y=1-(3/t)# at the point when t=2?

Please see the explanation for the process.

From the chain rule, we know that:

Evaluate x and y at t = 2:

Use the point-slope form of the equation of a line:

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To find the equation of the tangent line to the curve given by parametric equations ( x = 1 + \frac{1}{t^2} ) and ( y = 1 - \frac{3}{t} ) at the point when ( t = 2 ), follow these steps:

- Substitute ( t = 2 ) into the parametric equations to find the coordinates of the point on the curve corresponding to ( t = 2 ).

[ x = 1 + \frac{1}{2^2} = 1 + \frac{1}{4} = \frac{5}{4} ] [ y = 1 - \frac{3}{2} = 1 - \frac{3}{2} = -\frac{1}{2} ]

So, the point on the curve when ( t = 2 ) is ( \left(\frac{5}{4}, -\frac{1}{2}\right) ).

- Find the derivatives of ( x ) and ( y ) with respect to ( t ), denoted as ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ), respectively.

[ \frac{dx}{dt} = \frac{d}{dt}\left(1 + \frac{1}{t^2}\right) = -\frac{2}{t^3} ] [ \frac{dy}{dt} = \frac{d}{dt}\left(1 - \frac{3}{t}\right) = \frac{3}{t^2} ]

- Evaluate ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ) at ( t = 2 ).

[ \frac{dx}{dt}\bigg|*{t=2} = -\frac{2}{2^3} = -\frac{1}{4} ]
[ \frac{dy}{dt}\bigg|*{t=2} = \frac{3}{2^2} = \frac{3}{4} ]

- Use the point-slope form of the equation of a line to find the equation of the tangent line. The slope of the tangent line is ( \frac{dy}{dx} ) evaluated at ( t = 2 ).

[ \frac{dy}{dx}\bigg|*{t=2} = \frac{\frac{dy}{dt}\bigg|*{t=2}}{\frac{dx}{dt}\bigg|_{t=2}} = \frac{\frac{3}{4}}{-\frac{1}{4}} = -3 ]

- Plug the slope and the point ( \left(\frac{5}{4}, -\frac{1}{2}\right) ) into the point-slope form to get the equation of the tangent line.

[ y - y_1 = m(x - x_1) ] [ y - \left(-\frac{1}{2}\right) = -3\left(x - \frac{5}{4}\right) ] [ y + \frac{1}{2} = -3x + \frac{15}{4} ] [ y = -3x + \frac{13}{4} ]

Therefore, the equation of the tangent line to the curve at the point ( (5/4, -1/2) ) when ( t = 2 ) is ( y = -3x + \frac{13}{4} ).

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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.

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.

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.

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