# Simple Unit Vector Problem Part 2

1. Jul 7, 2014

### hagobarcos

1. The problem statement, all variables and given/known data

This is almost exactly the same problem as my earlier post, however different equation and
point.

Find the unit vector parallel to the graph f(x) = tan(x) and the point (pi/4 , 1) , and the
unit vector perpendicular to the graph and point (pi/4 , 1 ).

2. Relevant equations

Alright so following some advice from earlier, f'(x) = sec(x)tan(x)
at the point (pi/4 , 1), the slope of the function is f'(pi/4) = (1/(sqrt(2))/2)*(1) = (2)/(sqrt(2))

3. The attempt at a solution

So therefore, the parallel vector should have a base of < sqrt(2) , 2 >
and since the magnitude is sqrt (2 + 4) = sqrt (6)
then the unit vector parallel would then be <sqrt(2/6) , 2/sqrt(6) >

AND
the perp. vector is based off of the negative reciprocal of the parallel slope, equals -sqrt(2)/2
perp vector = < 2, -sqrt(2) >

unit perp. vector = <2/sqrt(6), -sqrt(1/3) >

Attached is a photo.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jul 7, 2014

### Staff: Mentor

Since when is the derivative of tan(x) equal to sec(x)tan(x)?

Chet

3. Jul 7, 2014

### hagobarcos

Since my derivatives and integrals are starting to blur together ^.^

Clearly then, f'(x) = sec^2(x)
During which, when x = pi/2, f'(x) = 2,

Vector par =< 1i, 2j >
Vhat = <1/sqrt(5) , 2/sqrt(5) >

Vector perp. = <2i, -1j >
Vp hat = <2/sqrt(5) , -1/sqrt(5) >

Bamn, web assign accepts, thank you!!
:D

4. Jul 7, 2014

### LCKurtz

Don't mix notations like that. Either write < 1,2> or write i + 2j.

5. Jul 7, 2014

### hagobarcos

Hmmm. I did not know this. Alright so if you have a vector ai + bj, it can be stated as <a, b> OR ai + bj .

May have been making up my own thing there. Please disregard previous method of writing them up.

6. Jul 8, 2014

### Fredrik

Staff Emeritus
Right, because by definition, i=<1,0> and j=<0,1>. So <a,b> = <a,0>+<0,b> = a<1,0>+b<0,1> = ai+bj.