Join the IDNLearn.com community and start finding the answers you need today. Get step-by-step guidance for all your technical questions from our knowledgeable community members.
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]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.
