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) = ()

Answer 1

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).