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. 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.
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