Question

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

(b) Find \(r'(t)\).

(c) Sketch the position vector \(r(t)\) and the tangent vector \(r'(t)\) for the given value of\(t\).

34.\(r(t) = < {t^2},{t^3} > ,t = 1\)

Short answer

(a) The curve for given equation is

(b) The required equation of the tangent vector is \(r'(t) = < 2t,3{t^2} > \)

(c) So the required graph is

Step-by-step solution

(a) Step 1: Value table for t

To draw the curve for vector equation \(r(t) = < {t^2},{t^3} > \). Let us put some values for t and find the corresponding values for \({t^2},{t^3}\).

\(\begin{array}{l}t = - 4,\;\;\;\;{t^2} = 16,\;\;{t^3} = - 64\\t = - 3,\;\;\;\;{t^2} = 9,\;\;\;{t^3} = - 27\\t = - 2,\;\;\;{t^2} = 4,\;\;\;\;{t^3} = - 8\\t = - 1,\;\;\;{t^2} = 1,\;\;\;\;\;{t^3} = - 1\end{array}\)

\(\begin{array}{l}t = 0,\;\;\;\;{t^2} = 0,\;\;\;\;\;{t^3} = 0\\t = 1,\;\;\;\;{t^2} = 1,\;\;\;\;\;\;{t^3} = 1\\t = 2,\;\;\;{t^2} = 4,\;\;\;\;\;{t^3} = 8\\t = 3,\;\;\;{t^2} = 9,\;\;\;\;\;{t^3} = 27\\t = 4,\;\;\;{t^2} = 16,\;\;\;{t^3} = 64\end{array}\)

(a) Step 2: Plot of graph

The curve for given equation is

(b) Step 3: To differentiate the given vector equation

To find r’(t) differentiate the given vector equation component-wise we get

\(\begin{array}{l}r(t) = < {t^2},{t^3} > \\r'(t) = \left\langle {\frac{{d{t^2}}}{{dt}},\frac{{d{t^3}}}{{dt}}} \right\rangle \\r'(t) = \left\langle {2t,3{t^2}} \right\rangle \end{array}\)

(b) Step 4: The final decision

The required equation of the tangent vector is \(r'(t) = < 2t,3{t^2} > \)

(c) Step 5: Simplification

\(\begin{array}{l}r(t) = < {t^2},{t^3} > \\t = 1\end{array}\)

(c) Step 6: At t = 1

\(r(1) = < 1,1 > \)

The tangent vector is (1,1)

Find the tangent vector at t = 1 by differentiating the given equation component wise and then put t = 1 that is

\(\begin{array}{l}r'(t) = < 2t,3{t^2} > \\r'(1) = < 2,3 > \end{array}\)

Draw the tangent vector \(r'(1)\) starting at \((1,1)\)going to \((1 + 2,1 + 3) = (3,4)\)