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What is the rectangular form of the parametric equations?

x(t)=t−2,y(t)=2t^2+4, where  t  is on the interval  [−3,1].

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y =[                  ] , where x is on the interval [        ,         ].

 May 15, 2020

Best Answer 

 #1
avatar+9015 
+2

x  =  t - 2

y  =  2t2 + 4

 

Let's solve the first equation for  t  and substitute that in for  t  into the second equation.

 

x  =  t - 2     Add  2  to both sides of this equation.

x + 2  =  t

 

Since  t  =  x + 2  we can substitute   x + 2  in for  t  into the second equation.

 

y  =  2t2 + 4

y  =  2(x + 2)2 + 4

 

I'll let you expand the right side of that equation if necessary.

 

To find the smallest possible value of  x,  plug in the smallest possible value of  t  into the equation   x = t - 2

And to find the biggest possible value of  x, plug in the biggest possible value of  t  into the equation   x = t - 2

 

When   t  =  -3,     x   =   -3 - 2   =   -5

When   t  =   1,     x   =    1 - 2   =   -1

 

So  x  is in the interval  [-5, -1]

 

Here's a graph:  https://www.desmos.com/calculator/ajymj9t4ve

 May 15, 2020
 #1
avatar+9015 
+2
Best Answer

x  =  t - 2

y  =  2t2 + 4

 

Let's solve the first equation for  t  and substitute that in for  t  into the second equation.

 

x  =  t - 2     Add  2  to both sides of this equation.

x + 2  =  t

 

Since  t  =  x + 2  we can substitute   x + 2  in for  t  into the second equation.

 

y  =  2t2 + 4

y  =  2(x + 2)2 + 4

 

I'll let you expand the right side of that equation if necessary.

 

To find the smallest possible value of  x,  plug in the smallest possible value of  t  into the equation   x = t - 2

And to find the biggest possible value of  x, plug in the biggest possible value of  t  into the equation   x = t - 2

 

When   t  =  -3,     x   =   -3 - 2   =   -5

When   t  =   1,     x   =    1 - 2   =   -1

 

So  x  is in the interval  [-5, -1]

 

Here's a graph:  https://www.desmos.com/calculator/ajymj9t4ve

hectictar May 15, 2020

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