Question

Determine the point at which the function r(t)= has a tangent line that is perpendicular to the plane 4x+y-2x=12

Answer 1

what is our r' ?

Answer 2

Wouldn't it be t^2, t, and 2t?

Answer 3

y'=4 but yeah

Answer 4

now what do we know about perp vectors and a dot product?

Answer 5

over thought that ... the normal to the plane is already perp to it, so we just equate it to the normal

Answer 6

Yeah I was about to say, don't they intersect?

Answer 7

<t^2, 4, 2t> = n<4,1,-2> assuming thats a -2z for the plane

Answer 8

Okay so now that we have that can we take the dot product?

Answer 9

dot product would give us a vector parallel to the plane

Answer 10

So cross?

Answer 11

as is, perp to the plane is in the same direction as the normal

recall that given a vector (a,b,c) and all vectors (x,y,z) perp to it; then the dot product gives us the equation of the plane

ax + by + cz = 0

Answer 12

so if we equate this to some scaled version of (a,b,c) then the line is in the same direction as the normal vector and is perp to the plane

Answer 13

<4,1,-2> could be our (a,b,c)
?

Answer 14

yep

Answer 15

And we can put that into our plane?

Answer 16

Or use the points from the plane as our (x,y,z)?

Answer 17

no, if we equate our r' parts with a scaled version of our (a,b,c) parts, then we will find a t value tha twill define the point we want

Answer 18

we want the point (t^2,4,2t) to equal some scaled version of (4,1,-2)

since we need y=4, let n=4

t^2,4,2t = 4(4),1(4),-2(4)

therefore t^2 = 16 and 2t = -8 and 4=4

Answer 19

solve for t

Answer 20

t=1?

Answer 21

1^2 not= 16
-2(1) not = -8

so im going with no, its not t=1

Answer 22

2(1) not = -8 .... either way lol