Respuesta :
The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.
What is vector?
It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.
We have vectored function:
[tex]\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})[/tex]
Find its derivative:
[tex]\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})[/tex]
Now its magnitude:
[tex]\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}[/tex]
After simplifying:
[tex]\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}[/tex]
Now the unit tangent is:
[tex]\rm T(u) = \dfrac{R'(t)}{|R'(t)|}[/tex]
After dividing and simplifying, we get:
[tex]\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)[/tex]
Now, finding the derivative of T(u), we get:
[tex]\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})[/tex]
Now finding its magnitude:
[tex]\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)[/tex]
After simplifying, we get:
[tex]\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}[/tex]
Now for the normal vector:
Divide T'(u) and |T'(u)|
We get:
[tex]\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})[/tex]
Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.
Learn more about the vector here:
https://brainly.com/question/8607618
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