Question

(a) Sketch the plane curve with the given vector equation

(b) Find

(c) Sketch the position vectorand the tangent vectorfor the given value of t

Short answer

(a) Plane curve with the given vector equation

(b) The \(r'(t) = \cos ti + \sin tj\)

(c) Position vector and the Tangent vector for the given value of t

Step-by-step solution

Step 1: Solution of the plane curve with the given function

Plane curve with the given vector equation

\(\begin{array}{l}r(t) = (1 + \cos t)i + (2 + \sin t)j,t = \frac{\pi }{6}\\r\left( {\frac{\pi }{6}} \right) = \left( {1 + \cos \left( {\frac{\pi }{6}} \right)} \right)i + \left( {2 + \sin \left( {\frac{\pi }{6}} \right)} \right)j\\r\left( {\frac{\pi }{6}} \right) = i + \frac{{\sqrt 3 i}}{2} + \frac{{5j}}{2}\end{array}\)

Step 2: Solution of 

The derivative ofis

\(\begin{array}{l}r(t) = (1 + \cos t)i + (2 + \sin t)j,t = \frac{\pi }{6}\\r'(t) = \int {(1 + \cos t)i + (2 + \sin t)j} \\r'(t) = \cos ti + \sin tj\end{array}\)

Step 3: Solution of position vector  and Tangent vector

Position vectorand the Tangent vectorfor the given value of t

\(\begin{array}{l}r(t) = (1 + \cos t)i + (2 + \sin t)j,t = \frac{\pi }{6}\\r\left( {\frac{\pi }{6}} \right) = \left( {1 + \cos \left( {\frac{\pi }{6}} \right)} \right)i + \left( {2 + \sin \left( {\frac{\pi }{6}} \right)} \right)j\\r\left( {\frac{\pi }{6}} \right) = i + \frac{{\sqrt 3 i}}{2} + \frac{{5j}}{2}\\r'(t) = \int {(1 + \cos t)i + (2 + \sin t)j} \\r'(t) = \cos ti + \sin tj\\r'\left( {\frac{\pi }{6}} \right) = - \frac{i}{2} + \frac{{j\sqrt 3 }}{2}\end{array}\)