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) = ()

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).