Oh! We can use that! r^2 = x^2+y^2 = (1-2*(y^2 / (x^2 + y^2)))^2, drop it into desmos and bam! Cartesian expression for a polar curve! Maybe you can clean that up a bit and simplify things, but that's just flavor: the core relationship between x and y is fully captured by this expression, and it's the same relationship captured by r=cos(2θ) Well, if it were an r^2, it would have to be equal to (1-2*(y^2 / (x^2 + y^2)))^2. Wait! We do! That square on the right hand side is to the full (y/r) quotient, so that's a r^2 in the denominator! Whoopie, we can capture the same r=cos(2θ) relationship by writing r=1-2*(y^2 / (x^2 + y^2))! Gah, still have an r on the left side. We could convert an r^2 into x^2+y^2, but we don't have an r^2. There's still r in a couple places, though. (*I didn't actually remember - I have it written down on my desk because I use it just infrequently enough to not have it memorized). Okay but does that help us re-write r=cos(2θ) with only x and y variables? You bet! Lots of different approaches possible here, but I happen to remember* that cos(2θ)=1-2sin(θ)^2. Oh! And since r is the distance from (0,0) to (x,y), we can say r^2=x^2+y^2! Similarly, we can use the height of the triangle to write down sin(θ)=y/r, and tan(θ)=y/x. So how can we keep the r=cos(2θ) relationship, but express it with x and y variables?Įasy! The base of this triangle is definitely x units long, which means cos(θ)=x/r. In a cartesian graph, you want the variables to be x: horizontal signed distance and y: vertical signed distance. For a polar graph, those variables are r: signed distance from the origin and θ: angle from horizontal. Every graph comes from a relationship between your variables. I see in line 8 you had the polar equation for a 4-leaved rose: r=cos(2θ). Here's a teaser: īut put that in the back of your mind for a moment, because I want to think about converting between polar and cartesian coordinates. That said, you probably learned about shifting (cartesian) curves in some earlier class! If you have a function f(x)=x^2 and you want to plot y=f(x+2), how should you move the original function? What about the graph for y+3=f(x)? (It may be helpful to think of it as y=f(x)-3) There's a lot of beautiful connections ready for you when you think about "adding or subtracting from the x- and y-coordinates" for graphs. Consider \(r=5 \cos \theta\) the maximum distance between the curve and the pole is \(5\) units.Hi! Looks like this was a challenge screen in your activity? My best guess is your teacher is thinking about this lesson as an early part of your investigation into polar equations, or maybe into plotting polar equations in Desmos specifically - they probably aren't expecting you to be fully proficient with the full mathematics here just yet. Set \(r=0\), and solve for \(\theta\).įor many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of \(\theta\) into the equation that result in the maximum value of the trigonometric functions. We use the same process for polar equations. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for \(x\). To find the zeros of a polar equation, we solve for the values of \(\theta\) that result in \(r=0\).
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |