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]
Learn more about the Parametric equation here:
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