Finding speed from position equation

[TehDarkArchon]

Homework Statement

We just had a test in my calc 3 class, and I'm pretty sure my teacher has the wrong solution to one of the answers. The question is about finding the speed of a particle given a space curve function r(t) = (cos2t)i + (3t - 1)j + (sin2t)k.

Homework Equations

v(t) = (x',y',z')
|v(t)| = √(x' + y' + z')

The Attempt at a Solution

For velocity I get (-2sin2t)i + (3)j + (2cos2t)k. Now, for speed I put √(4sin²2t + 4cos²2t + 9) = 2(sin2t + cos2t) + 3. He marked it wrong saying that it can be reduced to √(4+9) = √(13). I was under the impression that the sin²x + cos²x = 1 identity only works if the variable stands alone in the function, as sin2t/cos2t calls for a double angle formula.

 

[SammyS]

$\sqrt{a^2+b^2+c^2}\ne a+b+c$

Under the radical, you had $4\sin^22t + 4\cos^22t + 9$

This is equal to $4(\sin^22t + \cos^22t) + 9\ =\ 4(1)+9$

Look like your professor was right.

 

[TehDarkArchon]

Hmm I really never knew you could have any coefficient in front of x/t/whatever and still use the sin²x + cos²x = 1 identity. Oh well thanks a lot for the input

 

[Mark44]

I was under the impression that the sinx + cosx = 1 identity only works if the variable stands alone in the function, as sin2t/cos2t calls for a double angle formula.

I assume that you really meant sin²x + cos²x = 1.

If you have something like 9sin²(4x) + 9cos²(4x), you can factor out the 9 to get
9(sin²(4x) + cos²(4x)) = 9 * 1 = 9.

You don't have to expand sin(2x) or sin(4x) or whatever. The identity is sin²<whatever> + cos²<whatever> = 1, as long as the <whatever> is the same in both places.

 

[TehDarkArchon]

Ahh okay thank you both. That's very helpful for the future, and I guess my ignorance here is due to the fact that I've never really worked with a function like that before, but upon reviewing the derivation for the identity it does make sense. I feel pretty stupid for asking something so trivial, but thanks again