+69 The graph of the parametric equations 2 x=t and y=t+1 is shown to the right. Determine whether the graph would change for each set of parametric equations. If so, how would it change?
a) x=sin^2(t), y=sin(t)+1
b) x=t^4, y=t^2+1
>>>> Diagram: http://i.imgur.com/p6GRAdV.png <<<<
+118667 Thanks Chris,
What graphing program did you use Chris?
I played with this question too and I found it quite interesting. My results were the same as yours ofcourse. :)
Here are the Desmos graphs,
https://www.desmos.com/calculator/zcpydgu9sb
\(If \quad x=t^2\;\quad y=t+1\qquad \mbox{The restriction is } x\ge0\\ t=\pm\sqrt{x}\qquad t=y-1\\ y-1=\pm \sqrt{x}\\ (y-1)^2=x\\ x=(y-1)^2\\ \)
As Chris said this is a sidways parabola opening in the positive direction with a vertex at (0,1)
Lets look at the second one
a) x=sin^2(t), y=sin(t)+1
The first thing I notice is that t is replaced with sin(t)
so we have
\(x=sin^2t \qquad 0\le x\le 1\\ y=sin(t)+1 \qquad 0\le y \le 2\\ \mbox{So the domain and range are different.}\\ But\;\\ sin^{-1}(\pm\sqrt{x})=t \qquad and \qquad sin^{-1}(y-1)=t\\ \therefore\\ sin^{-1}(\pm\sqrt{x})=sin^{-1}(y-1)\\ \pm\sqrt{x}=(y-1)\\ x=(y-1)^2\\ x=(y-1)^2\\ \mbox{So this is exactly the same as the original}\\ \mbox{ relation except it has different a domain and range} \)
+129847 The first set of equations is
x = t^2 and y = t + 1
Re-arranging the second one, we have y - 1 = t and substituting this one into the first one, we have :
x = (y - 1)^2 which is a "sideways" parabola opening to the right with a vertex at (0, 1)
a) x = sin^2t y = sint + 1
Rearranging the second one again, we have sint = y - 1 and substituting this one into the first we have
x = (y - 1)^2
However......this graph is a "restricted" version of the first one......from t = 0 to t = pi/2, the graph traces the same path from the vertex (0,1) to the point (1, 2).....then from t = pi/2 to t =pi.....it "reverses" course from (1,2) back to the vertex......from t = pi to t = 3pi/2, it traces the same path as the original graph from the vertex to the point (1, 0)...lastly, from t = 3pi/2 to 2pi......the graph again reverses course and ends back at the vertex.......this behavior is repeated infinitly
Here's the graph :
b) x= t^4 y = t^2 + 1
Rearranging the second equation, we have y - 1 = t^2
Substituting this into the first equation we have the original equation, again, x = (y - 1)^2
However, this graph is just the [unbounded] upper half of the parabola since y is never less than 1
For the interval where t ranges from (-infinity, 0), the graph traces back from the right to the point (0, 1).....when t ranges from (0, infinity), the gragh "reverses" course and traces back to the right
Here's the graph here traced from t = -2 to t = 2 :
+118667 Thanks Chris,
What graphing program did you use Chris?
I played with this question too and I found it quite interesting. My results were the same as yours ofcourse. :)
Here are the Desmos graphs,
https://www.desmos.com/calculator/zcpydgu9sb
\(If \quad x=t^2\;\quad y=t+1\qquad \mbox{The restriction is } x\ge0\\ t=\pm\sqrt{x}\qquad t=y-1\\ y-1=\pm \sqrt{x}\\ (y-1)^2=x\\ x=(y-1)^2\\ \)
As Chris said this is a sidways parabola opening in the positive direction with a vertex at (0,1)
Lets look at the second one
a) x=sin^2(t), y=sin(t)+1
The first thing I notice is that t is replaced with sin(t)
so we have
\(x=sin^2t \qquad 0\le x\le 1\\ y=sin(t)+1 \qquad 0\le y \le 2\\ \mbox{So the domain and range are different.}\\ But\;\\ sin^{-1}(\pm\sqrt{x})=t \qquad and \qquad sin^{-1}(y-1)=t\\ \therefore\\ sin^{-1}(\pm\sqrt{x})=sin^{-1}(y-1)\\ \pm\sqrt{x}=(y-1)\\ x=(y-1)^2\\ x=(y-1)^2\\ \mbox{So this is exactly the same as the original}\\ \mbox{ relation except it has different a domain and range} \)
