# The surface #z=xsqrt(x+y)# intersects the plane y=3 along a curve c, how do you find the parametric equations for the tangent line to this curve at the point P(1,3,2)?

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The parametric equations for the tangent line to the curve ( c ) at the point ( P(1,3,2) ) can be found by first finding the partial derivatives of ( z ) with respect to ( x ) and ( y ), then evaluating them at the point ( P ) to find the direction vector of the tangent line.

Given the surface equation ( z = x\sqrt{x+y} ) and the plane equation ( y = 3 ), the curve ( c ) is the intersection of the surface and the plane.

Partial derivatives: [ \frac{\partial z}{\partial x} = \frac{\sqrt{x+y}}{2\sqrt{x}} + \sqrt{x+y} ] [ \frac{\partial z}{\partial y} = \frac{x}{2\sqrt{x(x+y)}} ]

Evaluate at ( P(1,3,2) ): [ \frac{\partial z}{\partial x} = \frac{\sqrt{3}}{2} + \sqrt{3} = \frac{3\sqrt{3}}{2} ] [ \frac{\partial z}{\partial y} = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6} ]

Direction vector: [ \mathbf{v} = \left(\frac{3\sqrt{3}}{2}, \frac{\sqrt{3}}{6}, 1\right) ]

Parametric equations for the tangent line: [ x(t) = 1 + \frac{3\sqrt{3}}{2}t ] [ y(t) = 3 + \frac{\sqrt{3}}{6}t ] [ z(t) = 2 + t ]

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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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