Respuesta :
Answer:
[tex]\left(-\frac 14, \frac 1{\sqrt 2}-\ln\left(\sqrt 2\right)\right)[/tex]
Step-by-step explanation:
We find the leftmost point when [tex]\frac{dx}{dt}=0[/tex]
[tex]\therefore 4t^3-2t=0[/tex]
[tex]\Rightarrow 2t(2t^2-1)=0[/tex]
[tex]\Rightarrow t=0, t=\frac 1{\sqrt 2}\; \text{and }\; t=-\frac 1{\sqrt 2}[/tex]
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 [tex]t=\frac 1{\sqrt 2}[/tex] ,
[tex]x=\left(\frac 1{\sqrt 2}\right)^4 -\left(\frac 1{\sqrt 2}\right)^2=\frac 14-\frac 12[/tex]
[tex]\Rightarrow x=-\frac 14[/tex]
[tex]y=\frac 1{\sqrt 2}+\ln\left(\frac 1{\sqrt 2}\right)[/tex]
[tex]\Rightarrow y=\frac 1{\sqrt 2}+\ln 1-\ln\left(\sqrt 2\right)[/tex]
[tex]\Rightarrow y=\frac 1{\sqrt 2}-\ln\left(\sqrt 2\right)[/tex]
Hence, the exact Cartesian coordinate of the leftmost point is: [tex]\left(-\frac 14, \frac 1{\sqrt 2}-\ln\left(\sqrt 2\right)\right)[/tex].
The leftmost point of the parametric curve
[tex]x=t^4-t^2\\y=t+ln(t)[/tex]
is
[tex](x,y)=(-\frac{1}{4},\frac{1}{\sqrt2}-\frac{1}{2}ln(2))[/tex]
Given the parametric curve
[tex]x=t^4-t^2\\y=t+ln(t)[/tex]
a leftmost point occurs when the following condition is satisfied
[tex]\frac{dx}{dt}=0,\frac{dy}{dt}\ne0[/tex]
and
[tex]\frac{d^2x}{dt^2}>0[/tex]
For the above curve
[tex]x=t^4-t^2\\\\\frac{dx}{dt}=4t^3-2t[/tex]
when [tex]\frac{dx}{dt}=0[/tex]
[tex]4t^3-2t=0\\2t(t\sqrt{2}-1)(t\sqrt{2}+1)=0\\\\t=0\text{ or }t=\frac{1}{\sqrt2}\text{ or }t=-\frac{1}{\sqrt2}[/tex]
because of [tex]y=t+ln(t)[/tex], we can only use [tex]t=\frac{1}{\sqrt2}[/tex]
But we have to test if [tex]\frac{dy}{dt}\ne0[/tex] at [tex]t=\frac{1}{\sqrt2}[/tex]
[tex]\frac{dy}{dt}|_{t=\frac{1}{\sqrt2}}\\\\={1+\frac{1}{t}}|_{t=\frac{1}{\sqrt2}}\\\\=1+\sqrt2\ne0[/tex]
To test if [tex]t=\frac{1}{\sqrt2}[/tex] gives the leftmost point,
[tex]\frac{d^2x}{dt^2}=12t^2-2\\\\[/tex]
at [tex]t=\frac{1}{\sqrt2}[/tex]
[tex]\frac{d^2x}{dt^2}=12(\frac{1}{\sqrt2})^2-2\\\\=4>0[/tex]
So, at [tex]t=\frac{1}{\sqrt2}[/tex], we have a leftmost point. The coordinates of the leftmost point are
[tex]x=t^4-t^2\\\\=(\frac{1}{\sqrt2})^4-(\frac{1}{\sqrt2})^2\\\\=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}[/tex]
and
[tex]y=t+ln(t)\\\\=(\frac{1}{\sqrt2})+ln(\frac{1}{\sqrt2})\\\\=\frac{1}{\sqrt2}-\frac{1}{2}ln(2)[/tex]
So we have a leftmost point at [tex]t=\frac{1}{\sqrt2}[/tex] which is
[tex](x,y)=(-\frac{1}{4},\frac{1}{\sqrt2}-\frac{1}{2}ln(2))[/tex]
Learn more about parametric functions here: https://brainly.com/question/10196044
