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zepdrixBest ResponseYou've already chosen the best response.2
Let's get it into standard form.\[\large \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1\] Hmm, so the x has a 2 on it. Multiplying by 2 is the same as dividing by 1/2.\[\large 2x^2 \qquad \rightarrow \qquad \frac{x^2}{\left(\frac{1}{2}\right)}\]Then let's apply the square to both the top and bottom term.\[\large \frac{x^2}{\left(\frac{1}{2}\right)} \qquad \rightarrow \qquad \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2\]Now we can clearly see what our \(a\) value is.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Doing similar to the y gives us something like this I believe, \[\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=1\] And then to graph it.. ummm trying to remember. If x=0, y extends out to \(\pm \sqrt2\). if y=0, x extends out to \(\pm \dfrac{1}{\sqrt2}\)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Hmm lemme throw this equation into a grapher real quick. I wanna make sure I didn't miss anything.
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
thank you, very well put explanation ... can you graph these with a ti83 or anything like that? and if that equation was set equal to 2 instead of 1, you would just set each variable = to zero and then solve for your new x and y
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=2\]Divide both sides by 2,\[\large \left(\frac{x}{\frac{\sqrt2}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2\cdot \sqrt2}\right)^2=1\]Which simplifies to,\[\large \left(\frac{x}{1}\right)^2+\left(\frac{y}{2}\right)^2=1\]These would be our new \(a\) and \(b\) values.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Calculator? :o hmm thinking.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
If you wanted to put it into your calculator, you would probably want to avoid the fancy ellipse standard form. From the start of the problem, \[\large 2x^2 + (y^2)/2 = 1\]Solving for y gives us,\[\large y=\pm 2\sqrt{12x^2}\]And you'd probably have to put it in as 2 separate functions,\[\large y_1=2\sqrt{12x^2} \qquad \qquad y_2=2\sqrt{12x^2}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
There's probably another way to do it, but I'm not too familiar with calculators :\
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
so we always want it in the form of =1 then... the next problem i got was x + ln y = C how would we graph that as C changes?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
If we solved for y, we would see that this is an exponential function.\[\large \ln y=C+x\]Exponentiating both sides,\[\large e^{\ln y}=e^{C+x} \qquad \rightarrow \qquad y=e^Ce^x\] As C changes? :O hmm it looks like it's just changing the scaling on our function. It just grows a lot faster as C increases.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
We can't graph an unknown constant :D I guess you could just graph a few different functions and compare. With different coeeficients in front, e^2, e^3 and so on.
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
these problems ask me to "graph several members of each family", and i assume they mean as C changes right
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
like just pick random C's and graph ??
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
When C=0, we have the graph \(y=e^x.\) When C=1, we have the graph \(y=e\cdot e^x\), so this one will grow about 3 times as fast as e^x. When C=2, we have, \(y=e^2\cdot e^x\), so this one will grow like 7 times faster than e^x. Yah just plug in random C's. Graph a few members of the family! :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Oh it's not just growing faster, it's also moving the yintercept :O I didn't think about that.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
https://www.desmos.com/calculator/2saf2toes8
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
does the members also include the original equation do you think?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
The equation we're graphing `is` the original equation, we've just moved some things around so it's easier to graph.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
https://www.desmos.com/calculator/agvkcwjylj
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
For example... the first one... the problem asks us to find the orthogonal trajectories for the family of curves and graph several members of each family. x = Cy^4 was the original problem and after we find the orthogonal trajectories our equation becomes 2x^2 + y^2/2 = C so I am unsure if I should just graph random C's of the resulting equation or both.
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
or what "several members of each family" actually means
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
I'm not exactly sure about the orthogonal trajectories, But here are several members of each family graphed, The equations should be displayed on the left. https://www.desmos.com/calculator/mqoozlfnq2
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
I guessing that's probably what they want you to do, to graph a few of those.
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
That is a cool looking graphing calculator... but i see that both equations are orthogonal... so I guess we did the problem right... so I would just graph both then since they probably want to see them orthogonal
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Oh I didn't notice that! Yah they intersect at right angles! Hah! That's cool :D
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
can anyone use that calculator... thats just on a website for everyone to use?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Yah I can't remember who showed it to me. Maybe Satellite or someone.. But I had to bookmark it immediately, very useful tool! :) Yah it's just a free resource I think.
 one year ago

zonazooBest ResponseYou've already chosen the best response.0
well thank you again. for all your help.
 one year ago
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