Question

Use calculus to find the exact Cartesian coordinates of the leftmost point on the parametric curvex=t4−t2, y=t+ln(t).(x,y)=(_____ , _____ )

Answer 1

Answer:

$\left(-\frac 14, \frac 1{\sqrt 2}-\ln\left(\sqrt 2\right)\right)$

Step-by-step explanation:

We find the leftmost point when  $\frac{dx}{dt}=0$

$\therefore 4t^3-2t=0$

$\Rightarrow 2t(2t^2-1)=0$

$\Rightarrow t=0, t=\frac 1{\sqrt 2}\; \text{and }\; t=-\frac 1{\sqrt 2}$

Now we not take the solutions 0 and -1/√2  because we take the leftmost point.

Thus, we cannot do in of anything less than or equal to 0.

For  $t=\frac 1{\sqrt 2}$ ,

$x=\left(\frac 1{\sqrt 2}\right)^4 -\left(\frac 1{\sqrt 2}\right)^2=\frac 14-\frac 12$

$\Rightarrow x=-\frac 14$

$y=\frac 1{\sqrt 2}+\ln\left(\frac 1{\sqrt 2}\right)$

$\Rightarrow y=\frac 1{\sqrt 2}+\ln 1-\ln\left(\sqrt 2\right)$

$\Rightarrow y=\frac 1{\sqrt 2}-\ln\left(\sqrt 2\right)$

Hence, the exact Cartesian coordinate of the leftmost point is: $\left(-\frac 14, \frac 1{\sqrt 2}-\ln\left(\sqrt 2\right)\right)$.