Question

eliminate the parameter to find Cartesian equation of the curve.

Answer 1

\[x=t^3+t\]
\[y= t^2+2\]

Answer 2

-2 </ t </ 2

Answer 3

t x y

-2 -10 6
-1 -2 3
0 0 2
1 2 3
2 10 6

Answer 4

brb gonna refresh open study is lagging me

Answer 5

\(\color{black}{x=t^3+t}\)
\(\color{black}{y=t^2+2\quad \Longleftrightarrow \quad y-2=t^2}\)

\(\color{black}{x^2=(t^3+t)^2}\)
\(\color{black}{x^2=t^6+2t^4+t^2}\)
\(\color{black}{x^2=(t^2)^3+2(t^2)^2+(t^2)}\)
\(\color{black}{x^2=(y-2)^3+2(y-2)^2+y-2}\)

Answer 6

Side question: Have you learned derivatives?

Answer 7

\(\color{black}{-2\le t \le 2}\)

\(\color{black}{\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{d}{dt}(t^2+2)}{\frac{d}{dt}(t^3+t)}= \frac{2t}{3t^2+1} }\) This is the slope of the curve.

So, the critical numbers are:
\(\color{black}{t=-2,0,+2}\) (The 2's are endpoints, 0 is a solution to dy/dx=0)

\(\color{black}{t=-2\rightarrow y=(-2)^2+(-2)=2,~x=(-2)^3+(-2)=-10\rightarrow (x,y)=(-10,2)}\)
\(\color{black}{t=0~~\rightarrow~~ y=0^2+0=0,~~x=0^3+0=0~~\rightarrow~~ (x,y)=(0,0)}\)
\(\color{black}{t=2~~\rightarrow~~ y=2^2+2=6,~~x=2^3+2=10~~\rightarrow~~ (x,y)=(10,6)}\)

So, the absolute minimum is (0,0) and the absolute maximum is at (10,6).

Answer 8

So, your domain is \(\color{black}{0\le x\le 10}\)

Answer 9

okay question

are we suppose to solve for a certain value?

Answer 10

I just solved for the absolute maximum and absolute minimum of your parametric function, to determine the domain.

Answer 11

how do you know which function to solve ?

and how did my teacher get