Q. 21 Let C be the graph of a vector-v... [FREE SOLUTION]

Chapter 11: Q. 21 (page 880)

Let C be the graph of a vector-valued function r. The plane determined by the vectors T(t₀) and B(t₀) and containing the point r(t₀) is called the rectifying plane for C at r(t₀). Find the equation of the rectifying plane to the helix determined by $r (t) = (\cos (t), \sin (t), t)$ when t = 蟺.

Short Answer

Expert verified

$x + 1 = 0$

Step by step solution

Step1. Given Information

$Consider r (t) = 鉄� \cos 鈦� t, \sin 鈦� t, t 鉄� at t = \frac{蟺}{2} The objective is to find the equations of the rectifying plane to r (t) at t = 蟺 . For this, T (t), N (t) and B (t) should be calculated first. Consider \begin{array}{l} r (t) = 鉄� \cos 鈦� t, \sin 鈦� t, t 鉄� \\ r^{鈥�} (t) = 鉄� 鈭� \sin 鈦� t, \cos 鈦� t, 1 鉄� \\ \begin{aligned} 鈭�r^{鈥�} (t)鈭� = \sqrt{\sin^{2} 鈦� t + \cos^{2} 鈦� t + 1} \\ = \sqrt{2} \end{aligned} \\ T (t) = \frac{r^{鈥�} (t)}{鈭�r^{鈥�} (t)鈭�} \\ = \frac{鉄� 鈭� \sin 鈦� t, \cos 鈦� t, 1 鉄�}{\sqrt{2}} \\ At t = 蟺, T (t) = T (蟺) = \frac{鉄� 鈭� \sin 鈦� 蟺, \cos 鈦� 蟺, 1 鉄�}{\sqrt{2}} \\ = \frac{鉄� 0, 鈭� 1, 1 鉄�}{\sqrt{2}} \\ = <0, \frac{鈭� \sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \\ Thus the unit tangent vector to r (t) at t = 蟺 is \\ \begin{array}{l} T (蟺) = <0, \frac{鈭� \sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \\ T^{鈥�} (t) = \frac{1}{\sqrt{2}} 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄� \\ 鈭�T^{鈥�} (t)鈭� = \sqrt{\frac{1}{2} \cos^{2} 鈦� t + \frac{1}{2} \sin^{2} 鈦� t} \\ = \frac{1}{\sqrt{2}} \\ N (t) = \frac{T^{鈥�} (t)}{鈭�T^{鈥�} (t)鈭�} = \frac{\frac{1}{\sqrt{2}} 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄�}{\frac{鈭� 1}{\sqrt{2}}} \\ = 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄� \\ At t = 蟺, N (t) = N (蟺) = 鉄� 鈭� \cos 鈦� 蟺, 鈭� \sin 鈦� 蟺, 0 鉄� \\ = 鉄� 1, 0, 0 鉄� \\ Thus the principal unit normal vector to r (t) at t = 蟺 is 鉄� 1, 0, 0 鉄� \\ B (蟺) = T (蟺) 脳 N (蟺) \\ \begin{array}{l} = |\begin{array}{ccc} i & j & k \\ 0 & \frac{鈭� \sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 1 & 0 & 0 \end{array}| \\ = i (0) 鈭� j (\frac{鈭� \sqrt{2}}{2}) + k (\frac{\sqrt{2}}{2}) \\ = <0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \end{array} \end{array} \end{array}$ uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('3012ea183b02056...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage(' $" width="0" height="0" role="math"> Consider r (t) = 鉄� \cos 鈦� t, \sin 鈦� t, t 鉄� at t = \frac{蟺}{2} The objective is to find the equations of the rectifying plane to r (t) at t = 蟺 . For this, T (t), N (t) and B (t) should be calculated first. r (t) = <e^{t}, e^{鈭� t}, \sqrt{2} t) at t = 0 Consider \begin{array}{l} r (t) = 鉄� \cos 鈦� t, \sin 鈦� t, t 鉄� \\ r^{鈥�} (t) = 鉄� 鈭� \sin 鈦� t, \cos 鈦� t, 1 鉄� \\ \begin{aligned} 鈭�r^{鈥�} (t)鈭� = \sqrt{\sin^{2} 鈦� t + \cos^{2} 鈦� t + 1} \\ = \sqrt{2} \end{aligned} \\ T (t) = \frac{r^{鈥�} (t)}{鈭�r^{鈥�} (t)鈭�} \\ = \frac{鉄� 鈭� \sin 鈦� t, \cos 鈦� t, 1 鉄�}{\sqrt{2}} \\ At t = 蟺, T (t) = T (蟺) = \frac{鉄� 鈭� \sin 鈦� 蟺, \cos 鈦� 蟺, 1 鉄�}{\sqrt{2}} \\ = \frac{鉄� 0, 鈭� 1, 1 鉄�}{\sqrt{2}} \\ = <0, \frac{鈭� \sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \\ Thus the unit tangent vector to r (t) at t = 蟺 is \\ \begin{array}{l} T (蟺) = <0, \frac{鈭� \sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \\ T^{鈥�} (t) = \frac{1}{\sqrt{2}} 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄� \\ 鈭�T^{鈥�} (t)鈭� = \sqrt{\frac{1}{2} \cos^{2} 鈦� t + \frac{1}{2} \sin^{2} 鈦� t} \\ = \frac{1}{\sqrt{2}} \\ N (t) = \frac{T^{鈥�} (t)}{鈭�T^{鈥�} (t)鈭�} = \frac{\frac{1}{\sqrt{2}} 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄�}{\frac{鈭� 1}{\sqrt{2}}} \\ = 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄� \\ At t = 蟺, N (t) = N (蟺) = 鉄� 鈭� \cos 鈦� 蟺, 鈭� \sin 鈦� 蟺, 0 鉄� \\ = 鉄� 1, 0, 0 鉄� \\ Thus the principal unit normal vector to r (t) at t = 蟺 is 鉄� 1, 0, 0 鉄� \\ B (蟺) = T (蟺) 脳 N (蟺) \\ \begin{array}{l} = |\begin{array}{ccc} i & j & k \\ 0 & \frac{鈭� \sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 1 & 0 & 0 \end{array}| \\ = i (0) 鈭� j (\frac{鈭� \sqrt{2}}{2}) + k (\frac{\sqrt{2}}{2}) \\ = <0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \end{array} \end{array} \end{array}$ uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('c83cb95b9d1d76b...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage(' $" width="0" height="0" role="math"> Consider r (t) = 鉄� \cos 鈦� t, \sin 鈦� t, t 鉄� at t = \frac{蟺}{2} The objective is to find the equations of the rectifying plane to r (t) at t = 蟺 . For this, T (t), N (t) and B (t) should be calculated first. r (t) = <e^{t}, e^{鈭� t}, \sqrt{2} t) at t = 0 Consider \begin{array}{l} r (t) = 鉄� \cos 鈦� t, \sin 鈦� t, t 鉄� \\ r^{鈥�} (t) = 鉄� 鈭� \sin 鈦� t, \cos 鈦� t, 1 鉄� \\ \begin{aligned} 鈭�r^{鈥�} (t)鈭� = \sqrt{\sin^{2} 鈦� t + \cos^{2} 鈦� t + 1} \\ = \sqrt{2} \end{aligned} \\ T (t) = \frac{r^{鈥�} (t)}{鈭�r^{鈥�} (t)鈭�} \\ = \frac{鉄� 鈭� \sin 鈦� t, \cos 鈦� t, 1 鉄�}{\sqrt{2}} \\ At t = 蟺, T (t) = T (蟺) = \frac{鉄� 鈭� \sin 鈦� 蟺, \cos 鈦� 蟺, 1 鉄�}{\sqrt{2}} \\ = \frac{鉄� 0, 鈭� 1, 1 鉄�}{\sqrt{2}} \\ = <0, \frac{鈭� \sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \end{array}$

$Thus the unit tangent vector to r (t) at t = 蟺 is \begin{array}{l} T (蟺) = <0, \frac{鈭� \sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \\ T^{鈥�} (t) = \frac{1}{\sqrt{2}} 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄� \\ 鈭�T^{鈥�} (t)鈭� = \sqrt{\frac{1}{2} \cos^{2} 鈦� t + \frac{1}{2} \sin^{2} 鈦� t} \\ = \frac{1}{\sqrt{2}} \\ N (t) = \frac{T^{鈥�} (t)}{鈭�T^{鈥�} (t)鈭�} = \frac{\frac{1}{\sqrt{2}} 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄�}{\frac{鈭� 1}{\sqrt{2}}} \\ = 鉄� 鈭� \cos 鈦� t, 鈭� \sin 鈦� t, 0 鉄� \\ At t = 蟺, N (t) = N (蟺) = 鉄� 鈭� \cos 鈦� 蟺, 鈭� \sin 鈦� 蟺, 0 鉄� \\ = 鉄� 1, 0, 0 鉄� \\ Thus the principal unit normal vector to r (t) at t = 蟺 is 鉄� 1, 0, 0 鉄� \\ B (蟺) = T (蟺) 脳 N (蟺) \\ \begin{array}{l} = |\begin{array}{ccc} i & j & k \\ 0 & \frac{鈭� \sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 1 & 0 & 0 \end{array}| \\ = i (0) 鈭� j (\frac{鈭� \sqrt{2}}{2}) + k (\frac{\sqrt{2}}{2}) \\ = <0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}> \end{array} \end{array}$

Step2. Rectifying plane

$The plane determined by the vectors T (t_{0}) and B (t_{0}) containing the point r (t_{0}) is called the rectifying plane for C at r (t_{0}) . Thus the equation of the rectifying plane to r (t) at t = 蟺 is (T (蟺) 脳 B (蟺)) 鈰� 鉄� x 鈭� x (蟺), y 鈭� y (蟺), z 鈭� z (蟺) 鉄� = 0 First computing T (蟺) 脳 B (蟺) :$

$\begin{aligned} = |\begin{array}{ccc} i & j & k \\ 0 & \frac{鈭� \sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array}| \\ = |\begin{array}{ccc} i & j & k \\ 0 & \frac{鈭� \sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array}| \end{aligned}$

$Evaluating \begin{array}{l} (T (蟺) 脳 B (蟺)) 鈰� 鉄� x 鈭� x (蟺), y 鈭� y (蟺), z 鈭� z (蟺) 鉄� = 0 \\ 鉄� 鈭� 1, 0, 0 鉄� 鈰� 鉄� x + 1, y, z 鈭� 蟺鉄� = 0 \\ 鈬� 鈭� 1 (x + 1) = 0 \\ 鈬� x + 1 = 0 \\ Thus the equation of the rectifying plane to r (t) at t = 蟺 is \\ x + 1 = 0 \end{array}$

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Step2. Rectifying plane

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Let C be the graph of a vector-valued function r. The plane determined by the vectors T(t0) and B(t0) and containing the point r(t0) is called the rectifying plane for C at r(t0). Find the equation of the rectifying plane to the helix determined by r(t)=(cost,sint,t)when t = 蟺.

Short Answer

Step by step solution

Step1. Given Information

Step2. Rectifying plane

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Statistics

Applied Mathematics

Mechanics Maths

Calculus

Decision Maths

Study anywhere. Anytime. Across all devices.