Mathematics: 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).

The paraboloid z = 6 − x − x2 − 5y2 intersects

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20.01.2022

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:

$z = 6-x-x^{2} -5y^{2}$

if x = 2, then

$z = 6-(2)-2^{2}-5y^{2}$

$z = -5y^{2}$

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

$z = -5t^{2}$ since $z = -5y^{2}$

We calculate the derivative

$\frac{dx}{dt}= 0$

$\frac{dy}{dt}= 1$

$\frac{dz}{dt}= -10t$

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

Thus:

$\frac{dx}{dt}= 0$

$\frac{dy}{dt}= 1$

$\frac{dz}{dt}= -10(2)= -20$

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

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