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Vika [28.1K]
3 years ago
15

Slope of a line perpendicular to 5x -9y = 1

Mathematics
1 answer:
Oksana_A [137]3 years ago
8 0
I think it would be y=(-9/5)x -1/9
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Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)
garik1379 [7]

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.

<h3>What is vector?</h3>

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:

\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})

Find its derivative:

\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})

Now its magnitude:

\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}

After simplifying:

\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}

Now the unit tangent is:

\rm T(u) = \dfrac{R'(t)}{|R'(t)|}

After dividing and simplifying, we get:

\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)

Now, finding the derivative of T(u), we get:

\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})

Now finding its magnitude:

\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)

After simplifying, we get:

\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t},          \sqrt{2}e^{9t},  \sqrt{2}e^{9t})

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:

brainly.com/question/8607618

#SPJ4

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