# How do you find the equation of the tangent to the curve #x=t^4+1#, #y=t^3+t# at the point where #t=-1# ?

So we use the derivatives of the parametric equations:

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To find the equation of the tangent to the curve ( x = t^4 + 1 ) and ( y = t^3 + t ) at the point where ( t = -1 ), you first find the coordinates of the point of tangency by substituting ( t = -1 ) into the equations for ( x ) and ( y ). This gives ( x = (-1)^4 + 1 = 2 ) and ( y = (-1)^3 + (-1) = -2 ).

Next, you find the derivative of ( y ) with respect to ( x ), ( \frac{dy}{dx} ), by differentiating ( y ) with respect to ( t ) and dividing by the derivative of ( x ) with respect to ( t ), ( \frac{dy}{dt} \div \frac{dx}{dt} ).

[ \frac{dy}{dt} = 3t^2 + 1 ] [ \frac{dx}{dt} = 4t^3 ]

Now, substitute ( t = -1 ) into these derivatives: [ \frac{dy}{dt} = 3(-1)^2 + 1 = 4 ] [ \frac{dx}{dt} = 4(-1)^3 = -4 ]

Then, find the slope of the tangent line by evaluating ( \frac{dy}{dx} ) at ( t = -1 ): [ \frac{dy}{dx} = \frac{4}{-4} = -1 ]

Finally, use the point-slope form of the equation of a line ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is the point of tangency and ( m ) is the slope: [ y - (-2) = -1(x - 2) ]

Simplify to get the equation of the tangent line: [ y + 2 = -x + 2 ] [ y = -x ]

Therefore, the equation of the tangent to the curve at the point where ( t = -1 ) is ( y = -x ).

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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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