Finding the Normal Plane to a Curve at a Given Point

[Septcanmat]

Homework Statement

At what point on the curve ⃗r(t) = (t^3, 3t, t^4) is the normal plane parallel to the plane 3x + 3y − 4z = 9 (the normal plane is the plane through the point ⃗r(t) which is normal to ⃗r′(t))

Homework Equations

I'm not really sure.

The Attempt at a Solution

(6t)(x-t^3) + (0)(y-3t) + (8t)(z-t^4) = 0

But that got me nowhere.


Thanks in advance.

 

[lanedance]

i'm not sure what you attempted there...

first find vector normal to the plane given, then find the tangent vector of the curve.. and have a think about how they will be related

 

[Septcanmat]

I read somewhere that that would be the equation of a normal plane to a curve. But it didn't work.

I did what you said, and they'll be related in that they'll be parallel vectors. I tried doing what you said and setting them equal to each other, but I just got equations for t that seem insolvable. ( for instance, 0= 8t^4 + 9 +16t^6)

 

[lanedance]

that's not what i get, it works out ok... I'm not sure how you get the higher powers of t in your equation either

what do you get for the normal to the plane & for the tangent vector?

 

[Septcanmat]

The normal is the gradient, so I got (3,3,-4). And I got (3t^2,3,4t^3)/sqrt(9t^4+9+16t^6) for the tangent vector.

 

[lanedance]

both look good, but I see you are normalising the tangent vector to length 1, that's not needed here, as you just need to know when its parallel

so if p is normal to the plane, t is the tangent, you just need to know when t = c.p for any constant c, which shows they are parallel. This should lead to a reasonably easy equation set if you don't normalise the vector

 

[Septcanmat]

Well alright then, lol. Thanks guys, got it all figured out now. The answer is (-1,-3,1). Or rather, I'm assuming that that's the correct answer because it's what I got and it matches up with one of the multiple choice options :P

 

[lanedance]

yep that's what i get