logistic_guy
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Sketch.
\(\displaystyle \bold{r}(t) = (t + 1)\bold{i} + (t^2 - 2)\bold{\bold{j}}\)
\(\displaystyle \bold{r}(t) = (t + 1)\bold{i} + (t^2 - 2)\bold{\bold{j}}\)
sketch
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logistic_guy
Senior Member
- Joined
- Apr 17, 2024
- Messages
- 2,214
We can write this curve in terms of \(\displaystyle x\) and \(\displaystyle y\) if we are smart enough. Then, it is a piece of cake to recognize its graph!
\(\displaystyle x = t + 1\)
This means:
\(\displaystyle t = x - 1\)
This gives:
\(\displaystyle y = t^2 - 2 = (x - 1)^2 - 2\)
This is just the graph of a parabola that is shifted one unit to the right and two units downward and concaves upward with its vertex at \(\displaystyle (1,-2)\)
Since this is a parametrized curve, we expect it contains a path. It should be counterclockwise in this case. Why? Think about it!
\(\displaystyle x = t + 1\)
This means:
\(\displaystyle t = x - 1\)
This gives:
\(\displaystyle y = t^2 - 2 = (x - 1)^2 - 2\)
This is just the graph of a parabola that is shifted one unit to the right and two units downward and concaves upward with its vertex at \(\displaystyle (1,-2)\)
Since this is a parametrized curve, we expect it contains a path. It should be counterclockwise in this case. Why? Think about it!