Find the slope and angle of inclination

[gillgill]

1) Find the slope
a) (3.14)/6 b) 135° c)60°

2) Find the angle of inclination of a line with the given slope
a) m=-1/2 b) m=1 c) m=-2 d) m=57

how do u do these questions?

 

[xanthym]

1) Find the slope
a) (3.14)/6 b) 135° c)60°

2) Find the angle of inclination of a line with the given slope
a) m=-1/2 b) m=1 c) m=-2 d) m=57

how do u do these questions?

1) {Slope} = tan{Angle of Inclination}
2) {Angle of Inclination} = tan^(-1){Slope}

Notes:
a) tan^(-1) = arctan = {Inverse tan}
b) In #1 above, determine the units of each "Angle of Inclination" before doing calculations and be sure calculator is set to the proper units (e.g., degrees or radians).


~~

 

[Kamataat]

Hello, gillgill, I'll do a few for you:

The slope of a line passing through points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is given by $$m=y_2-y_1/x_2-x_1 = \tan\alpha$$, where $$\alpha$$ is the angle between the line and the x-axis.

So, if given an angle $$\alpha=\pi/6=180^o/6=30^o$$, you find the slope by evaluating $$m=\tan30^o$$, which I believe is $$m=\sqrt{3}/3$$.

Now, if given a slope $$m=-1/2$$, you know that $$m=-1/2=\tan\alpha$$. If you know the value of $$\tan\alpha$$ (in this case $$-1/2$$), then you can find $$\alpha$$, by evaluating $$\arctan m=\arctan(-1/2)=\alpha$$, which gives you an angle of inclination $$\alpha=-26.57^o$$.

Hope this helps! Now try to do the other ones by yourself.

- Kamataat

 

[gillgill]

ok..icic...
but for ex. 1a) the answer is 1/√3...how do u find that out?

 

[gillgill]

o..okok..i see now...thx guys

 

[Kamataat]

$$1/\sqrt{3}=\sqrt{3}/3$$

- Kamataat

 

[gillgill]

how do u go from tan30 to 1/√3?

 

[ek]

how do u go from tan30 to 1/√3?

Things you should know:

sin 30 = cos60 = 1/2
cos 30 = sin 60 = √3/2
tan 30 = sin30/cos30 = 1/2 / √3/2 = 1/√3

You should also know that sin45 and cos45 are both √2/2. And therefore tan45 is 1.

It is also helpful to remember approximate values for √3/2 and √2/2. They are .866 and .707 respectively. I can remember in high school my knowledge of those approximations helped me answer many questions I wasn't exactly sure of, quite quickly.

 

[gillgill]

ic...
thanks

 

[dextercioby]

If u're to remember someting,try to remember $\sqrt{2}\approx 1.414$ and $\sqrt{3}\approx 1.732$ or maybe with 3 sign.figures,only...

That way,u can do whatever operations with them.

Daniel.

 

[gillgill]

i have a different question this time...
given y=√x-4
reflected in the y-axis
translated 5 units left
compressed horizontally by 1/3

i keep getting y=√-3(x+1)...which is wrong...can anybody tell me what my mistake is?

 

[primarygun]

How can I type latex faster?>

 

[HallsofIvy]

i have a different question this time...
given y=√x-4
reflected in the y-axis
translated 5 units left
compressed horizontally by 1/3

i keep getting y=√-3(x+1)...which is wrong...can anybody tell me what my mistake is?

How in the world did you get that negative??

In general, a horizontal (left-right) change is a change in x.

Translating 5 units left shifts means that "x= -5" acts like "x= 0": that is the formula must involve "x+ 5". (Shifting 0 to a is the same as replacing x by x-a: in this case a= -5.)

"Compressing horizontally by 1/3" means "x= 1/3" acts like "x=1". That will be true if we multiply x by 3: replace x by 3x.

Putting those together, we can "translate 5 units left" and then "compress horizontally by 1/3" by replacing x by 3(x+ 5).

Since the original function was y= √(x-4) (I am assuming the entire x-4 was inside the root), the new function will be √(3(x+5)-4)=
√(15x+ 11).

 

[gillgill]

does reflect in the y-axis gives x an negative?
by the way...the answer is y=√-3(x+3)

 

[Kamataat]

This is what I did:

Original function: $$y=\sqrt{x-4}$$

Reflect on the y-axis: $$y=\sqrt{-x-4}$$

Translate horizontally 5 units to the left: $$y=\sqrt{-(x+5)-4}=\sqrt{-x-5-4}=\sqrt{-x-9}$$

Compress horizontally by 1/3: $$y=\sqrt{-3x-9}=\sqrt{-3(x+3)}$$

- Kamataat

 

[gillgill]

o..i get it now...thx