Question

Find equations of the normal plane and osculating plane of the curve at the given point.

\(x = 2\sin 3t,y = t,z = 2\cos 3t\); \(\left( {0,\pi , - 2} \right)\)

Short answer

The equation of normal plane and osculating plane of curve with parametric equation \(x = 2\sin 3t\) , \(y = t\) , and \(z = 2\cos 3t\) at points \(\left( {0,\pi , - 2} \right)\) are \(y = \pi + 6x\) and \(x + 6y = 6\pi \) respectively.

Step-by-step solution

Formula used

1. The vector function with parametric equations\(x = f\left( t \right)\),\(y = g\left( t \right)\), and\(z = h\left( t \right)\)is\(r\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \).
2. The tangent vector of a vector function\(r\left( t \right)\)is\(T\left( t \right) = \frac{{r'\left( t \right)}}{{\left| {r'\left( t \right)} \right|}}\).
3. The normal vector of a vector function\(r\left( t \right)\)is\(N\left( t \right) = \frac{{T'\left( t \right)}}{{\left| {T'\left( t \right)} \right|}}\).
4. The Bi-normal vector of vector function\(r\left( t \right)\)is\(B\left( t \right) = T\left( t \right) \times N\left( t \right)\).
5. The equation of plane with vector \(r'\left( t \right) = \left\langle {{r_1},{r_2},{r_3}} \right\rangle \) at point \(\left( {{x_1},{y_1},{z_1}} \right)\) is \({r_1}\left( {x - {x_1}} \right) + {r_2}\left( {y - {y_1}} \right) + {r_3}\left( {z - {z_1}} \right) = 0\) .

Calculation

The given parametric equations of plane are\(x = 2\sin 3t\),\(y = t\), and\(z = 2\cos 3t\)and point\(\left( {0,\pi , - 2} \right)\).

Then, by formula(1) the vector equation with given parametric equations is\(r\left( t \right) = \left\langle {2\sin 3t,t,2\cos 3t} \right\rangle \).

Equate the components of\(r\left( t \right)\)with point\(\left( {0,\pi , - 2} \right)\)as follows.

\(\begin{aligned}{l}2\sin 3t = 0\\t = \pi \\2\cos 3t = - 2\end{aligned}\)

Hence, the value of\(t\)is\(\pi \).

Compute the first derivative of\(r\left( t \right) = \left\langle {2\sin 3t,t,2\cos 3t} \right\rangle \)with respect to\(t\)on both sides.

Calculate the derivative value\(r'\left( t \right)\)at point\(t = \pi \).

\(\begin{aligned}{l}r'\left( \pi \right) = \left\langle {6\cos 3\left( \pi \right),1, - 6\sin 3\left( \pi \right)} \right\rangle \\ = \left\langle {6\cos 3\pi ,1, - \sin 3\pi } \right\rangle \\ = \left\langle {6\left( { - 1} \right),1,6\left( 0 \right)} \right\rangle \\ = \left\langle { - 6,1,0} \right\rangle \end{aligned}\)

Apply the formula(5) and obtain the equation of normal plane of vector\(r'\left( \pi \right) = \left\langle { - 6,1,0} \right\rangle \)at the point\(\left( {0,\pi , - 2} \right)\).

\(\begin{aligned}{l} - 6\left( {x - 0} \right) + 1\left( {y - \pi } \right) + 0\left( {z + 2} \right) = 0\\ - 6x + y - \pi + 0 = 0\\y = \pi + 6x\end{aligned}\)

The magnitude of\(r'\left( t \right) = \left\langle {6\cos 3t,1, - 6\sin 3t} \right\rangle \)is obtained as follows.

Substitute\(r'\left( t \right) = \left\langle {6\cos 3t,1, - 6\sin 3t} \right\rangle \)and\(\left| {r'\left( t \right)} \right| = \sqrt {37} \)in formula(2) to obtain the tangent vector of a vector function\(r\left( t \right)\).

\(\begin{aligned}{l}T\left( t \right) = \frac{{\left\langle {6\cos 3t,1, - 6\sin 3t} \right\rangle }}{{\sqrt {37} }}\\ = \left\langle {\frac{{6\cos 3t}}{{\sqrt {37} }},\frac{1}{{\sqrt {37} }},\frac{{ - 6\sin 3t}}{{\sqrt {37} }}} \right\rangle \end{aligned}\)

Compute the first derivative of\(T\left( t \right) = \left\langle {\frac{{6\cos 3t}}{{\sqrt {37} }},\frac{1}{{\sqrt {37} }},\frac{{ - 6\sin 3t}}{{\sqrt {37} }}} \right\rangle \)with respect to\(t\).

\(\begin{aligned}{l}T'\left( t \right) = \frac{d}{{dt}}\left\langle {\frac{{6\cos 3t}}{{\sqrt {37} }},\frac{1}{{\sqrt {37} }},\frac{{ - 6\sin 3t}}{{\sqrt {37} }}} \right\rangle \\ = \left\langle {\frac{d}{{dt}}\left( {\frac{{6\cos 3t}}{{\sqrt {37} }}} \right),\frac{d}{{dt}}\left( {\frac{1}{{\sqrt {37} }}} \right),\frac{d}{{dt}}\left( {\frac{{ - 6\sin 3t}}{{\sqrt {37} }}} \right)} \right\rangle \\ = \left\langle {\frac{{ - 18\sin 3t}}{{\sqrt {37} }},0,\frac{{ - 18\cos 3t}}{{\sqrt {37} }}} \right\rangle \end{aligned}\)

The magnitude of\(T'\left( t \right) = \left\langle {\frac{{ - 18\sin 3t}}{{\sqrt {37} }},0,\frac{{ - 18\cos 3t}}{{\sqrt {37} }}} \right\rangle \)is obtained as follows.

Substitute\(T'\left( t \right) = \left\langle {\frac{{ - 18\sin 3t}}{{\sqrt {37} }},0,\frac{{ - 18\cos 3t}}{{\sqrt {37} }}} \right\rangle \)and\(\left| {T'\left( t \right)} \right| = \frac{{18}}{{\sqrt {37} }}\)in formula(3) to obtain the normal vector of a vector function\(r\left( t \right)\).

\(\begin{aligned}{l}N\left( t \right) = \frac{{\left\langle {\frac{{ - 18\sin 3t}}{{\sqrt {37} }},0,\frac{{ - 18\cos 3t}}{{\sqrt {37} }}} \right\rangle }}{{\frac{{18}}{{\sqrt {37} }}}}\\ = \frac{{\sqrt {37} }}{{18}}\left\langle {\frac{{ - 18\sin 3t}}{{\sqrt {37} }},0,\frac{{ - 18\cos 3t}}{{\sqrt {37} }}} \right\rangle \\ = \left\langle { - \sin 3t,0, - \cos 3t} \right\rangle \end{aligned}\)

Note that, the cross product of\(u\left( t \right) = \left\langle {{u_1}\left( t \right),{u_2}\left( t \right),{u_3}\left( t \right)} \right\rangle \)and\(v\left( t \right) = \left\langle {{v_1}\left( t \right),{v_2}\left( t \right),{v_3}\left( t \right)} \right\rangle \)is given by;

\(\begin{aligned}{l}u\left( t \right) \times v\left( t \right) = \left| {\begin{aligned}{*{20}{c}}i&j&k\\{{u_1}\left( t \right)}&{{u_2}\left( t \right)}&{{u_3}\left( t \right)}\\{{v_1}\left( t \right)}&{{v_2}\left( t \right)}&{{v_3}\left( t \right)}\end{aligned}} \right|\\ = \left\langle {\left( {\left( {{u_2}\left( t \right){v_3}\left( t \right) - {v_2}\left( t \right){u_3}\left( t \right)} \right)} \right),\left( { - \left( {{u_1}\left( t \right){v_3}\left( t \right) - {v_1}\left( t \right){u_3}\left( t \right)} \right)} \right),\left( {\left( {{u_1}\left( t \right){v_2}\left( t \right) - {v_1}\left( t \right){u_2}\left( t \right)} \right)} \right)} \right\rangle \end{aligned}\)

Substitute\(T\left( t \right) = \left\langle {\frac{{6\cos 3t}}{{\sqrt {37} }},\frac{1}{{\sqrt {37} }},\frac{{ - 6\sin 3t}}{{\sqrt {37} }}} \right\rangle \)and\(N\left( t \right) = \left\langle { - \sin 3t,0, - \cos 3t} \right\rangle \)in the formula(4) to obtain bi-normal vector of vector function\(r\left( t \right)\).

\(B\left( t \right) = T\left( t \right) \times N\left( t \right)\)

\(\begin{aligned}{l} = \left\langle {\frac{{6\cos 3t}}{{\sqrt {37} }},\frac{1}{{\sqrt {37} }},\frac{{ - 6\sin 3t}}{{\sqrt {37} }}} \right\rangle \times \left\langle { - \sin 3t,0, - \cos 3t} \right\rangle \\\left( {\left( {\left( {\frac{1}{{\sqrt {37} }}} \right)\left( { - \cos 3t} \right) - \left( 0 \right)\left( {\frac{{ - 6\sin 3t}}{{\sqrt {37} }}} \right)} \right)} \right),\\ = \left\langle {\left( { - \left( {\left( {\frac{{6\cos 3t}}{{\sqrt {37} }}} \right)\left( { - \cos 3t} \right) - \left( { - \sin 3t} \right)\left( {\frac{{ - 6\sin 3t}}{{\sqrt {37} }}} \right)} \right)} \right),} \right\rangle \\\left( {\left( {\left( {\frac{{6\cos 3t}}{{\sqrt {37} }}} \right)\left( 0 \right) - \left( { - \sin 3t} \right)\left( {\frac{1}{{\sqrt {37} }}} \right)} \right)} \right)\\ = \left\langle {\frac{{ - \cos 3t}}{{\sqrt {37} }} - 0,\frac{{6{{\cos }^2}3t}}{{\sqrt {37} }} + \frac{{6{{\sin }^2}3t}}{{\sqrt {37} }},0 - \frac{{ - \sin 3t}}{{\sqrt {37} }}} \right\rangle \\ = \left\langle {\frac{{ - \cos 3t}}{{\sqrt {37} }},\frac{6}{{\sqrt {37} }},\frac{{ - \sin 3t}}{{\sqrt {37} }}} \right\rangle \end{aligned}\)

Calculate the derivative value\(B\left( t \right)\)at the point\(t = \pi \).

\(\begin{aligned}{l}B\left( \pi \right) = \left\langle {\frac{{ - \cos 3\left( \pi \right)}}{{\sqrt {37} }},\frac{6}{{\sqrt {37} }},\frac{{ - \sin 3\left( \pi \right)}}{{\sqrt {37} }}} \right\rangle \\ = \left\langle {\frac{{ - \left( { - 1} \right)}}{{\sqrt {37} }},\frac{6}{{\sqrt {37} }},\frac{0}{{\sqrt {37} }}} \right\rangle \\ = \left\langle {\frac{1}{{\sqrt {37} }},\frac{6}{{\sqrt {37} }},0} \right\rangle \end{aligned}\)

Find the equation of osculating plane of vector\(B\left( \pi \right) = \left\langle {\frac{1}{{\sqrt {37} }},\frac{6}{{\sqrt {37} }},0} \right\rangle \)at point\(\left( {0,\pi , - 2} \right)\)by using formula(5).

\(\begin{aligned}{l}\frac{1}{{\sqrt {37} }}\left( {x - 0} \right) + \frac{6}{{\sqrt {37} }}\left( {y - \pi } \right) + 0\left( {z + 2} \right) = 0\\\frac{1}{{\sqrt {37} }}x + \frac{6}{{\sqrt {37} }}y - \frac{6}{{\sqrt {37} }}\pi + 0 = 0\\\frac{x}{{\sqrt {37} }} + \frac{{6y}}{{\sqrt {37} }} = \frac{{6\pi }}{{\sqrt {37} }}\\x + 6y = 6\pi \end{aligned}\)

Thus, the equation of normal plane and osculating plane of curve with parametric equations \(x = 2\sin 3t\) , \(y = t\) , and \(z = 2\cos 3t\) at point \(\left( {0,\pi , - 2} \right)\) are \(y = \pi + 6x\) and \(x + 6y = 6\pi \) respectively.