calc help

[Pumpkin_Jack]

Well I figyre someone here at the lair.may be able to help me with thus homework question...

Prove the head of the vector function r(t)=tcosi + tsintj + tk lies on the cone z^2=x^2+y^2

I'm on my phone so this took forever to write but please someone help me I'm so lost in calc 3

[Pumpkin_Jack]

anyone? the bold letters are vectors

[Sirex]

42?

[Skud]

42?

Lmao!

[Pumpkin_Jack]

Y'all are no help. Neither is google.

[Dead...Again]

Sorry man, I haven't done complex math in 13 years.

If you had questions about C or ASM programming, I could hook you up!

[lady-MGS]

That is WAY above my head. Maybe this site can help? If you scroll down, there is a calculus section.

http://www.khanacademy.org/

[Pumpkin_Jack]

i figured it out, was going about it the wrong way what i needed to do was.....

Convert r(t) into <tcost, tsint, t> then use that as the plug ins for <x,y,z>
comming up with
t^2=(tcost)^2 + (tsint)^2
divide both sides by t^2 to get
1=(cost)^2 + (sint)^2 which is a property that shows that at the point i converted r(t) is the point at which both lines intercect because the property above is true.

[ChrisWW11]

I think using the derivative may have also helped.
r(t)=tcosi + tsintj + tk

d/dt(t*cos(i)+t*sin(tj)) = sin(jt)+j*t cos(jt)+cosh(1)