Find the equations of the normal plane and osculating plane of the curve r(t) = \langle e^t,...

Question:

Find the equations of the normal plane and osculating plane of the curve {eq}\mathbf r(t) = \langle e^t, e^t\sin t, e^t\cos t\rangle {/eq} at the point (1,0,1) which corresponds to {eq}t=0 {/eq}

Answer and Explanation:

Let the curve {eq}\vec{r}\left ( t \right ) = \left \langle e^{t}, e^{t}\sin t, e^{t} \cos t \right \rangle {/eq} at the given region {eq}\left ( 1, 0, 1 \right ) {/eq} at {eq}t = 0 {/eq}

And {eq}\left | r'\left ( t \right ) \right | = \sqrt{3} e^{t} {/eq}

Differentiate the given curve, we get

{eq}\vec{r'}\left ( t \right ) = \left \langle e^{t}, e^{t} \cos t + e^{t} \sin t, -e^{t}\sin t + \cos t e^{t} \right \rangle \\ = \left \langle e^{t}, e^{t}\left ( \cos t + \sin t \right ), e^{t}\left ( \cos t - \sin t \right ) \right \rangle {/eq}

When {eq}t = 0 \\ \vec{r'}\left ( t \right ) = \left \langle 1, 1, 1 \right \rangle {/eq}

Let's find the normal plane:

{eq}1\left ( x - 1 \right ) + 1\left ( y - 0 \right ) + 1\left ( z - 1 \right ) = 0 \\ x - 1 + y + z - 1 = 0 \\ x + y + x = 2 {/eq}


Learn more about this topic:

How to Solve a System of Equations by Substitution

from 6th-8th Grade Math: Practice & Review

Chapter 29 / Lesson 4
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