Slope field Question

[nanase]

Hello,
Can anybody explain the meaning of the question? I have sound knowledge in derivative and integral but fairly new to this slope field topic
my answer key says for (a) the answer is y=-x and for (b) the answer is y=-x+pi

I need guidance on how to reach those answers. What should i be looking at.
Cheers

 

[BigBeachBanana]

If you draw in the given answers, notice that the line [imath]y=\pi-x[/imath] intersects the relative maxima of both solutions, while [imath]y=-x[/imath] intercepts the relative minimum of both solutions. This is what they're asking for.

Now, the question is how do you find the equation of these lines?
Recall from calculus, [imath]\frac{dy}{dx}=0[/imath] will give you the critical points. Then take the domains into consideration.

 

[The Highlander]

I need guidance on how to reach those answers. What should i be looking at.
Cheers

For maxima & minima \(\displaystyle \frac{dy}{dx}\) must be zero.

For \(\displaystyle \frac{dy}{dx}\) to be zero then \(\displaystyle sin(x+y) = 0 \Rightarrow (x+y) = 0 \space\text{or}\space \pi. \space \) (\(\displaystyle 2\pi\), etc. are outside range.)

So, \(\displaystyle y=-x \space\text{or}\space y=-x+\pi\)

Now draw up tables of possible candidate points and see which have horizontal slope(s) in the field to check.
(Domain of y means you need not consider negative values for it.)

For L₁:-
Do slopes in field appear to be horizontal at any of these points?
Certainly seem to at (0, 0) & (-2, 2), so y=-x is looking good!
y-intercept (C) occurs when x=0 so find C for your L₁ equation & Bob's your uncle! ?

When \(\displaystyle \mathbf{y=-x}\)

(Green Tick at those points where slope(s) appear horizontal in the field.)​

Continue for L₂ (When \(\displaystyle \mathbf{y=-x+\pi}\)) ...

 

[nanase]

May I ask why 2 pi is outside of range? where do you see it or how can we conclude it?
.
.
do we look at the y-values? because maximum of y is 5, while 2 pi is more than 6?

 

[The Highlander]

May I ask why 2 pi is outside of range? where do you see it or how can we conclude it?
.
do we look at the y-values? because maximum of y is 5, while 2 pi is more than 6?

\(\displaystyle Sin(x+y)=0 \Rightarrow (x+y)=\space..., -2 \pi, - \pi,  0,  \pi,  2\pi,...\)

So,  \(\displaystyle y=-x  \text{or}  y=k\pi-x\)

Graphs of: \(\displaystyle  y=-x  \text{and}  y=k\pi-x, \text{for}  k=-\pi  \text{to}  2\pi  (\pi  \text{steps})\) illustrate that only \(\displaystyle \\y=-x  \text{and}  y=\pi-x\) pass through the maxima & minima of the graphs in your slope field (compare these plots to those provided by BBB, above; I couldn't superimpose them the way s/he did ?) all other values of k are even further away!

The Red box represents the Domain of x & y but, even though other lines pass through it, only the two meet the criteria specified (as stated above).

Hope that helps.

​

 

[BigBeachBanana]

 

[The Highlander]

Best I can do to combine the graphs; trust you get the idea? ?

​

 

[The Highlander]

Never mind! BBB beats me to it again! ?

 

[The Highlander]

Got it! Finally! ?
(Thanks to BBB's help. ?)

 

[nanase]

wow so it's not about the y values , but more on checking whether other lines will pass through the curves or not yeah?

Thank you for the illustrations, it really helps!!!
thanks BBB and the highlander (are they still immortal?) heheheh

 

[The Highlander]

wow so it's not about the y values , but more on checking whether other lines will pass through the curves or not yeah?

Thank you for the illustrations, it really helps!!!
thanks BBB and the highlander (are they still immortal?) heheheh

There can be only one! ?

But I'm sure there will also be a purely algebraic solution too, eg: BBB started by solving the Differential Equation (sec(x+y) − tan(x+y) = −x + C) that's how he got the graphs to come up (much) better than me. ?