Find the function r that satisfies the given condition. r'(t) = (e^t, sin t, sec^2 t): r(0) = (2, 2, 2) r(t) = ()

Question

Find the function r that satisfies the given condition. r'(t) = (e^t, sin t, sec^2 t): r(0) = (2, 2, 2) r(t) = ()

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Arianna 3 months 2021-10-19T22:28:13+00:00 1 Answer 0 views 0

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    2021-10-19T22:30:12+00:00

    Answer:

    r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2)

    Step-by-step explanation:

    A primitive of e^t is e^t+c, since r(0) has 2 in its first cooridnate, then

    e^0+c = 2

    1+c = 2

    c = 1

    Thus, the first coordinate of r(t) is e^t + 1.

    A primitive of sin(t) is -cos(t) + c (remember that the derivate of cos(t) is -sin(t)). SInce r(0) in its second coordinate is 2, then

    -cos(0)+c = 2

    -1+c = 2

    c = 3

    Therefore, in the second coordinate r(t) is equal to -cos(t)+3.

    Now, lets see the last coordinate.

    A primitive of sec²(t) is tan(t)+c (you can check this by derivating tan(t) = sin(t)/cos(t) using the divition rule and the property that cos²(t)+sin²(t) = 1 for all t). Since in its third coordinate r(0) is also 2, then we have that

    2 = tan(0)+c = sin(0)/cos(0) + c = 0/1 + c = 0

    Thus, c = 2

    As a consecuence, the third coordinate of r(t) is tan(t) + 2.

    As a result, r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2).

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