## Multivariable Calculus, 7th Edition

a) $y=(x+1)^{2}$ $x \gt -1$ b) As $t$ increases, $x$ also increases. So the arrow should be pointing in the direction in which x increases. There should be a hole at $(-1,0)$.
a) Given that $x=e^{t}-1$ $x+1=e^{t}$ Also given that $y=e^{2t}$ we can write as $y=(e^{t})^{2}$ Use the first equation to replace $e^{t}$ with $x+1$ $y=(x+1)^{2}$ (After graphing) Domain: $x \gt -1$ Range: $y \gt 0$ b) we know that $y=(x+1)^{2}$ where $x \gt -1$ Given that $x=e^{t}-1$ As $t$ increases, $x$ also increases. So the arrow should be pointing in the direction in which x increases. There should be a hole at $(-1,0)$.