Unit Vector

taurus · #1 $Old$ April 30th, 2008, 02:40 PM

f(x, y) = x(y^4)+y(x^4)

P(0, (1/(2*sqrt(2))))

Q(-1/(2*sqrt(2)), 0)

Whats the unit vector u in the direction from P to Q?
How do you do that?
Thanks

flyingsquirrel · #2 $Old$ April 30th, 2008, 03:02 PM

Hi

$\vec{PQ}$ is a vector which direction is from $P$ to $Q$ . (easy

) To build $\vec{u}$ from it, you only need to change its norm so that it equals 1. The solution is to do $\vec{u}=\frac{\vec{PQ}}{\|\vec{PQ}\|}$ because $\| \vec{u}\|=\left\|\frac{\vec{PQ}}{\|\vec{PQ}\|}\right\|=\frac{\|\vec{PQ}\|}{\|\vec{PQ}\|}=1$ .

taurus · #3 $Old$ April 30th, 2008, 10:17 PM

Huh am a bit confused what to do there? as i need an equation in the forum i j k as far as i know

earboth · #4 $Old$ April 30th, 2008, 11:30 PM

Quote:

Originally Posted by taurus $View Post$

f(x, y) = x(y^4)+y(x^4)

P(0, (1/(2*sqrt(2))))

Q(-1/(2*sqrt(2)), 0)

Whats the unit vector u in the direction from P to Q?
How do you do that?
Thanks

1. I don't know what the equation of the function has to do with your vector problem ...

2. Calculate the vector $\overrightarrow{PQ}=\overrightarrow{OQ} - \overrightarrow{OP}= \left(-\frac1{2 \sqrt{2}}~,~-\frac1{2 \sqrt{2}} \right)$

3. Now calculate the length (or value) of $\overrightarrow{PQ}$

$\left| \overrightarrow{PQ} \right|=\sqrt{\left( -\frac1{2 \sqrt{2}} \right)^2 + \left( -\frac1{2 \sqrt{2}} \right)^2}=\sqrt{\frac18 + \frac18}=\sqrt{\frac14} = \frac12$

4. To get the unit vector which has the same direction as $\overrightarrow{PQ}$ you have to divide the vector $\overrightarrow{PQ}$ by it's length (or value):

$\overrightarrow{PQ}^0 = \frac{\left(-\frac1{2 \sqrt{2}}~,~-\frac1{2 \sqrt{2}} \right)}{\frac12} = \left(-\frac1{ \sqrt{2}}~,~-\frac1{ \sqrt{2}} \right)$

5. In your notation you get:

$\overrightarrow{u_{PQ}} = -\frac1{\sqrt{2}} \cdot i + \left(-\frac1{\sqrt{2}}\right) \cdot j$

taurus · #5 $Old$ April 30th, 2008, 11:48 PM

Also this question based on the same data:
Finally use what you have found in the previous question parts to determine the directional derivative of f at P in the direction towards Q.

Now what i did is the gradient f * u
which gave me:
((y^4)+4*y*(x^3) + 4*x*(y^3) + (x^4)) * ((-1/sqrt(2))*i + (-1/sqrt(2))*j)

But that is wrong. Whats right?