# A question on vectors.

1. Jul 24, 2006

### MathematicalPhysicist

my question is from klppner's text an introduction to mechanics, on page 26, they suppose a vector A(t) has a constant magnitude A, he states that the direction dA(t)/dt is always perpendicular to A(t), now they define $$\triangle A=A(t+\triangle t)-A(t)$$, he states that $$|\triangle A|=2Asin(\frac{\triangle\theta}{2})$$ i dont understand how did they arrive at the last equation, and dont understand which angle does he refer to. he states that it's defined in a sketch in the book but i cant find where does he state this?

2. Jul 24, 2006

### neutrino

I don't have the book with me now, but I think $$\Delta\theta$$ is the angle between the $$A(t)$$ and $$A(t+\Delta t)$$. The equation becomes more accurate as the angle becomes smaller.

3. Jul 24, 2006

### quasar987

$\Delta \theta$ is the angle between$A(t+\triangle t)$ and A(t). Since both vectors are of magnitude A, the law of cosines gives

$$\Delta A^2 = A^2+A^2-2AAcos(\Delta \theta)$$

And $cos(2x) = 1 - 2sin^2(x)$, so...

4. Jul 24, 2006

### benorin

Ok, Draw a vector for A(t), and then draw another for $$A(t+\triangle t)$$ with the same length and starting from the same point in space. The angle $$\triangle\theta$$ is acute angle between these two vectors and $$\triangle A$$ is the vector that forms the third side of the isosceles triangle having A(t) and $$A(t+\triangle t)$$ for sides (and it terminates in the same place $$A(t+\triangle t)$$ does). Construct the altitude of this triangle and consider one of the resulting identical right triangles with one leg being the altitude, one leg is of length $$\frac{1}{2}\left| \triangle A\right|$$, and the hypotenuse is of length A. Use trig.

5. Jul 24, 2006

### Data

$$|\Delta A| = \sqrt{(A(t+\Delta t)-A(t))\cdot(A(t+\Delta t)-A(t))}$$

$$= \sqrt{|A(t+\Delta t)|^2 - 2A(t+\Delta t)\cdot A(t) + |A(t)|^2} = \sqrt{A^2 - 2A^2\cos {\Delta \theta} + A^2}$$

$$= \sqrt{2A^2 - 2A^2\cos{\Delta \theta}} = \sqrt{2}A\sqrt{1-\cos{\Delta \theta}}.$$

Now use the identity $\sin^2{\phi} = \frac{1}{2}(1-\cos{2\phi})$ to change that into

$$\sqrt{2}A\sqrt{2\sin^2{\frac{\Delta \theta}{2}}}$$

$$= 2A\sin \frac{\Delta \theta}{2}.$$

Here, $\Delta \theta$ is the angle between $A(t+\Delta t)$ and $A(t)$ and $A$ is the (constant) magnitude of the vector function.

6. Jul 24, 2006

### MathematicalPhysicist

7. Jul 24, 2006

### MathematicalPhysicist

by the way, neutrino, you havent bought the booog, have you?
ive seen in amazon that it's quite coasts a lot.

8. Jul 24, 2006

### neutrino

What's that? Some new variant for the word book?

9. Jul 25, 2006

### MathematicalPhysicist

no just big funny typing error.