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Im back! Um, i have a simple question im trying to get ready for test after 5 days.. I slacked of sadly :( on math, so i have to pick up my skills.. On my test review i have this question:

The parametric equations of a vector are $x_t=3+5t$ and $y_t=-1+3t$. Find the vector equation and the Cartesian equation.

I dont understand above, because i couldnt remember what my teacher told me about this one.. Can somone help me out! Much appreciated for reading thankyou!

amanuel2
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1 Answers1

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Presumably, the vector equation is the vector $r=\left[3+5t,-1+3t\right]$ or the equation $r=\left[5,3\right]t+\left[3,-1\right]$, and the Cartesian equation would just be $y=\frac{3}{5}(x-3)-1$ or $y=\frac{3}{5}x-\frac{14}{5}$. Just solve for the parameter $t$ in the first equation, and then substitute it into the second.

$t=\frac{1}{5}(x_t-3)$, and so $y_t = -1 + 3(\frac{1}{5}(x_t-3))$.

JAustin
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  • Hmm are you sure this is right anwser? Im getting skeptical :/ – amanuel2 May 25 '16 at 01:18
  • Yes, I believe so. You can plot the functions on WolframAlpha or plug in points if you're not sure. If you have questions about the method, feel free to ask. – JAustin May 25 '16 at 01:23
  • I dont know how to check either.. But i believe you :) – amanuel2 May 25 '16 at 01:27
  • You can make sure they're the same by plugging in values for $t$ or $x$ and $y$. We know $x_t=3+5t$ and $y_t=-1+3t$, so if $t$ is equal to $5$, for instance, $x$ is equal to $28$ and $y$ is equal to 14. Then try plugging $x=28$ into the Cartesian equation ($14=\frac{3}{5}28-\frac{14}{5}$). Do the same for a few different points until you're convinced. – JAustin May 25 '16 at 01:31
  • So what is the vector equation and cartesian equation? Sorry i didnt pay attention in class :( – amanuel2 May 25 '16 at 01:34
  • This is an important subject. A function of one variable (i.e. $f(x)$) maps on variable to another uniquely. For each value of $x$, there is a unique value of $y$. So for the function $y=x^2$, every value of x is associated with a single y value. $3^2 = 9, (3.5)^2=12.25$, etc. So the Cartesian function is a function that relates the basic Cartesian coordinates, $x, y$, and $z$.

    The same function can in general also be written as a parametric function of a parameter, $t$, which is mapped to not one but two values. Instead of mapping $x$ to $y$, we map $t$ to both $x$ and $y$.

     – JAustin May 25 '16 at 01:47
  • Instead of the function $y=x^2$, we have a vector $r$, which is equal to $\left[t,t^2\right]$. The $x$ coordinate is always equal to the t coordinate, and the $y$ coordinate is equal to the $t$ coordinate squared. The parameter $t$ is often used because the parameter often represents time. Instead of saying that the y coordinate of some particle is equal to the x coordinate squared, you can say that the particle is found at some $(x,y)$ coordinate at time $t$. – JAustin May 25 '16 at 01:51