Question

Ive gone through with someone on parts one and two on the question below, but I am stuck on the last part :

"are the graphs of: inverse tan(x) + inverse tan (y) = -pi/4 AND y=((x+1)/(x-1)) from above the same? EXPLAIN"

Can anyone help on that one?

I was thinking that the graphs would be the same since they are the same equation rewritten? But I was second guessing that because maybe they would look different rewritten in different forms?

ORIGINAL QUESTION BELOW:
show that the solution of:

dy/dx = -((1+y^2)/(1+x^2)) y(0) = -1
is the inverse tan (x) + inverse tangent (y) = -pi/4 AND show that the above can be written :
y= ((x+1)/(x-1))
And are the graphs of:
inverse tan(x) + inverse tan (y) = -pi/4 AND y=((x+1)/(x-1)) from above the same?

Explain.

Answer 1

have you implicitly found the derivative of arctan(x)+arctan(y)=-pi/4

Answer 2

Also is y(0)=-1 when arctan(x)+arctan(y)=-pi/4

Answer 3

in other words does this hold:
arctan(0)+arctan(-1)=-pi/4 ?

Answer 4

Yes I think i answered part one correctly, that inverse tanx + inverse tany = -pi/4, I am having a lot of trouble on the next two parts more so I guess

Answer 5

I have been trying to work backwards from the second derivative to the original function but it is not working out, am I on the right track?

Answer 6

why are you finding second derivative?

Answer 7

I would think all you need to do for the first part is confirm arctan(0)+arctan(-1)=-pi/4
and the implicity find the deirvative of arctan(x)+arctan(y)=-pi/4 and solve for dy/dx .
For the second part it sounds like we need to solve arctan(x)+arctan(y)=-pi/4 for y

Answer 8

\[\arctan(y)=\frac{-\pi}{4}-\arctan(x) \\ y=\tan(\frac{-\pi}{4}-\arctan(x)) \\ \text{ use difference identity for tan }\]

Answer 9

I meant first derivative, looking over it again, I tried to take a first derivative of y=(x+1)/(x-1) but I am not coming up with the original dy/dx function

Answer 10

I don't know if that even makes sense, I am so confused.

Answer 11

are you trying to do the second write now ?

Answer 12

second part?

Answer 13

Yes

Answer 14

then this is the way I would proceed
\[\arctan(y)=\frac{-\pi}{4}-\arctan(x) \\ y=\tan(\frac{-\pi}{4}-\arctan(x)) \\ \text{ use difference identity for \tan } \]

Answer 15

@Emilyjones85 is that going okay for you?

Answer 16

so that would be (tan(-pi/4) - tan (tan inverse (x)))/(1+tan (-pi/4) tan(inverse tan(x)) ?

Answer 17

yes y= but yes

Answer 18

yes sorry I forgot that

Answer 19

so tan(-pi/4) = -1 right?

Answer 20

do you know what tan(-pi/4)=?
and tan(arctan(x))=?

Answer 21

and yes on the first one

Answer 22

I will give you a hint tan and arctan are inverse functions

Answer 23

the domain of arctan(x) is actually all real numbers

Answer 24

so tan(arctan(x)) will always equal just x

Answer 25

i might be right back
i have to go do something real quick

Answer 26

right, so its (-1-x)/(1-x) - how does the numerator become x+1 though, am I missing something.

Answer 27

And on the last part would it just be that they are equal because they are the same thing?

Answer 28

\[y=\frac{-1-x}{1-x}=\frac{-1(1+x)}{1-x}=\frac{1+x}{(-1)(1-x)}\]

Answer 29

since -1/1=1/-1

Answer 30

Ok got it, sorry I had to get off last night. So that all makes sense now. And Like i was saying, the graphs would be the same, right? Because they turn out to be the same functions, right?