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).

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:

if x = 2, then

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  

 since

We calculate the derivative

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

Thus:

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 value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12