Acute Angle between two vector-valued functions

  • Thread starter Thread starter SolfegeDuck
  • Start date Start date
  • Tags Tags
    Angle Functions
SolfegeDuck
Messages
6
Reaction score
0

Homework Statement


Quick question. I need to find cos \vartheta, where \vartheta is the angle (acute) between two vector-valued functions (which are tangent lines).

Homework Equations



I think this is relevant:

cos\vartheta = (u dot v)/(\left\|u|| * \left\|v||)

The Attempt at a Solution



Is that the correct equation to use? If so, well, the two equations I'm using both have "t" in them, since they're VVFs. But what do I put in for t? I get the feeling that it's blatantly obvious...Thanks for helping!
 
Physics news on Phys.org
Yes, that's the right equation. Dot product is defined as multiplying i, j, and k components of each vector (assuming 3 component vectors) to return a scalar value. |u||v| means multiplying the scalar magnitudes of vector u and v.
 
Great, thanks! But when I actually do out the equation, I have all of those "t" variables everywhere (2t+1) - (3t^3), etcetera...what do I substitute in for that?
 
That depends on the question. Either it gives you a specific instance for when the two vectors cross or it wants the angle in general form where t is ambiguous.
 
The point is (1,1,3). But for u, t = 3, and for v, t = 4. So, do I use one of those t-variables, or go with something else entirely?
 
It shouldn't be hard to see that you need to use both of them- one applies to u and the other to v. In fact, instead of using t in both functions, it would be better to use, say, t1 for u and t2 for v. Then set t1= 3, t2= 4.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top