The paraboloid z = 6 − x − x2 − 5y2 intersects the plane x = 2 in a parabola. Find parametric equations in terms of t for the tangent line to this parabola at the point (2, 2, −20).

Respuesta :

Answer:

x = 2

y = 2 +  t

z = -20 -20t

Step-by-step explanation:

First, we are going to find the equation for this parabola. We replace x = 2 in the equation of the paraboloid, thus:

[tex]z = 6-x-x^{2} -5y^{2}[/tex]

if x = 2, then

[tex]z = 6-(2)-2^{2}-5y^{2}[/tex]

[tex]z = -5y^{2}[/tex]

Now, we calculate the tangent line to this parabola at the point (2,2,-20)

The parametrization of the parabola is:

x = 2

y = t  

[tex]z = -5t^{2}[/tex]  since [tex]z = -5y^{2}[/tex]

We calculate the derivative

[tex]\frac{dx}{dt}= 0[/tex]

[tex]\frac{dy}{dt}= 1[/tex]

[tex]\frac{dz}{dt}= -10t[/tex]

we evaluate the derivative in t=2, since at the point (2,2,-20) y = 2 and y = t

Thus:

[tex]\frac{dx}{dt}= 0[/tex]

[tex]\frac{dy}{dt}= 1[/tex]

[tex]\frac{dz}{dt}= -10(2)= -20[/tex]

Then, the director vector for the tangent line is (0,1,-20)

and the parametric equation for this line is:

x = 2

y = 2 +  t

z = -20 -20t

The parametric equation of the tangent line is [tex]L(t)=(2,2+t,-20-20t)[/tex]

Parabola :

The equation of Paraboloid is,

                 [tex]z =6-x-x^{2} -5y^{2}[/tex]

Equation of parabola when [tex]x = 2[/tex] is,

       [tex]z=6-2-2^{2} -5y^{2} \\\\z=-5y^{2}[/tex]

The parametric equation of parabola will be,

     [tex]r(t)=(2,t,-5t^{2} )[/tex]

Now, we have to find Tangent vector to this parabola is,

    [tex]T(t)=\frac{dr(t)}{dt}=(0,1,-10t)[/tex]

We get, the point [tex](2, 2, -20)[/tex] when [tex]t=2[/tex]

The tangent vector will be,

 [tex]T(2)=(0,1,-20)[/tex]

The tangent line to this parabola at the point (2, 2, −20) will be,

     [tex]L(t)=(2,2,-20)+t(0,1,-20)\\\\L(t)=(2,2+t,-20-20t)[/tex]

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