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Find the unit tangent vector at the given value of t for the following parameterized curves.

r(t)= ⟨7–√e^t,3e^t,3e^t⟩

, for 0≤t≤1
; t=ln 2

Sagot :

Answer:

[tex]T(ln2)[/tex][tex]= <\sqrt{145} , \frac{6\sqrt{2} }{\sqrt{145} } , \frac{6\sqrt{2} }{\sqrt{145} } >[/tex]

Step-by-step explanation:

Given : [tex]r(t)=<7-\sqrt{e^t}, 3e^t,3e^t > .........(i)[/tex]

We first have to differentiate of equation [tex](i)[/tex]

[tex]r'(t) = < \frac{\sqrt{e^t} }{2} , 3e^t,3e^t> ..........(ii)[/tex]

Now to get a unit tangent vector at the given value of [tex]t[/tex], we put [tex]t = ln2[/tex] in equation [tex](ii)[/tex]

[tex]r'(ln2) = < \frac{\sqrt{e^{ln2}} }{2} , 3e^{ln2},3e^{ln2}>[/tex]

            [tex]= <\frac{1}{\sqrt{2} } ,6,6>[/tex]        [∵[tex]e^{lna}=a[/tex]]

Now to get a unit tangent vector , we will divide our vector [tex]r'(ln2)[/tex] by its magnitude. So let's first find the magnitude.

[tex]|r'(ln2)|=\sqrt{(\frac{1}{\sqrt{2} } )^2 +6^2+6^2}[/tex]

             [tex]= \sqrt{\frac{1}{2}+36+36 }[/tex]

             [tex]= \sqrt{\frac{145}{2} }[/tex]

Now we can find the our unit tangents vector.

      [tex]T(ln2)=\frac{r'(ln2)}{|r'(ln2)|}[/tex]

                  [tex]= \frac{<\frac{1}{\sqrt{2}} ,6,6> }{\sqrt{\frac{145}{2}} }[/tex]

                  [tex]= <\frac{\frac{1}{\sqrt{2} } }{\sqrt{\frac{145}{2}} } ,\frac{6}{{\sqrt{\frac{145}{2}} }}, \frac{6}{{\sqrt{\frac{145}{2}} }} >[/tex]

                   [tex]= <\sqrt{145} , \frac{6\sqrt{2} }{\sqrt{145} } , \frac{6\sqrt{2} }{\sqrt{145} } >[/tex]

Hence, [tex]T(ln2)[/tex] [tex]= <\sqrt{145} , \frac{6\sqrt{2} }{\sqrt{145} } , \frac{6\sqrt{2} }{\sqrt{145} } >[/tex]

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