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Math Help - Calculus of Vector-Valued Func.

  1. #1
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    Calculus of Vector-Valued Func.

    1.) Given: r(t)=tcos(t)i+e(t^2)j+ln(t)k,

    what is r'(t)?

    2.) Suppose r(t) is a vector-valued func.,

    What geometric/graphical info does r'(a) tell us?

    3.) Let r(t)=cos(t)i+sin(t)j and s(t)=sin(5t)i+cos(5t)j.

    (a) What does the graphs of r(t) and s(t) look like?

    (b) Suppose the graphs of two vector-valued func. r(t) and s(t) are the same, then must r'(0)=s'(0)? (Explain whether this is a new result, or was it also true for functions f(x) and g(x)?)
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  2. #2
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    Hello, fifthrapiers!

    Here's some help . . .


    1) Given: . \vec{r}(t)\;=\;t\cos(t)\vec{i} + e^{t^2}\vec{j} + \ln(t)\vec{k}. . Find \vec{r'}(t)
    \vec{r'}(t)\;=\;(\cos t - t\sin t)\vec{i} + \left(2te^{t^2}\right)\vec{j} + \left(\frac{1}{t}\right)\vec{k}


    3) Let r(t)\:=\:\cos(t)i + \sin(t)j and s(t)\:=\:\sin(5t)i + \cos(5t)j

    (a) What does the graphs of r(t) and s(t) look like?
    Both are unit circles, centered at the origin.


    (b) Suppose the graphs of two vector-valued func. r(t) and s(t) are the same,
    . . .then must r'(0) = s'(0)? . no
    Consider: . \begin{array}{cc}\vec{r}(t) \;= & \cos(t)\vec{i} + \sin(t)\vec{j} \\ \vec{s}(t) \:= & \cos(t + \pi)\vec{i} + \sin(t + \pi)\vec{j} \end{array}


    When t = 0:

    . . \begin{array}{ccc}r(0) = \langle 1,\,0\rangle & \text{ and } & r'(0) = \langle 0,\,1\rangle\!:\;\uparrow \\ \\<br />
s(0) = \langle \text{-}1,0\rangle & \text{ and } &s'(0) = \langle 0,\,\text{-}1\rangle\!: \;\downarrow\end{array} . . . different derivatives



    Explain whether this is a new result,
    or was it also true for functions f(x) and g(x) ?

    This is a new result.

    With rectangular functions, y = f(x) and y = g(x),
    . . if their graphs are identical, then the functions are identical.
    . . Hence, their derivatives are equal.

    With parametric functions, a graph can be generated in a number of ways.
    . . As seen in part (b), their derivatives need not be equal.

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