Locate the open interval(s) on which the vectorvalued function's curve appears.r()=2cos3i+3sin3j,0≤≤2&nbsp;is smooth.

Question

Locate the open interval(s) on which the vectorvalued function&#039;s curve appears.r()=2cos3i+3sin3j,0≤≤2&amp;nbsp;is smooth.

Jick1984 · Accepted Answer

Step 1Consider the given vector-valued function:r(t)=(2cos3t)i^+(3sin3t)j^Here the objective is to find the open intervals on which the curve given by the vector-valued function is smooth.Let r(t)=f(t)i^+g(t)j^+h(t)k^ be the parameterization of the curve that is differentiable (i.e f′(t),g′(t) and h′(t) is continuous) on an open interval I. Then r(t) is smooth on the open interval I, if r′(t)≠0―. for any value of t in the interval I.Step 2The given vector-valued function isr(t)=(2cos3t)i^+(3sin3t)j^Differentiate r(t) with respect to tr′(t)=(−6sin2t)i^+(9cos2t)j^To find the open intervalr′(t)=0→{sin2⁡t=0cos2⁡t=0Here required such value of t for which sin2t=0 and cos2t=0 simultaneously.To solve the above equation, draw the graph of sin2t=0 and cos2t=0From the graph, it is clear that there is no such value of t for which sin2t=0 and cos2t=0 simultaneously.Therefore r′(t)≠0 ∈ (−∞,∞)Hence the curve is smooth on (−∞,∞)

Find the open interval(s) on which the curve given by

Answered question

Answer & Explanation

New Questions in Calculus 2