00:01
Hi there, this is a vector valid integral.
00:04
So let me just briefly explain what is a vector valid integral.
00:09
For example, if r of t is given as, let's say, f of t times i plus g of t times j plus h of d times k, so the integral of this function, so the integral of r of t, dt, which is also, equal to the integral of f of t d t and times i and plus integral of gt d t d t and plus integral of ht d t times k so this is the integral that we have to follow for this question as a step so first of all what am i supposed to i'm going to just try to find the r of t so i'm going to take the derivative of the derivative of the r of t so r of t which is equal to the integral of the derivative of the function and this can be written as e of t i plus sine t j and plus secant 2 t k so we can just take the derivative of this function and also i can write this function as separately which is e of t d t i and integral sine t d t j and integral zc and two t d t so this is the integral that we have to evaluate so the integral of e of t which is e of t d t let's say e of t c1 times i and the integral of which is negative cosine t so let me just put a negative here negative cosine plus c2 and j this is what we have here and plus the ccant 2 t so in finding the integral of this one so we have to just use the substitution rule so let's say let me just separate like and show secant 2t d t so for the integral of this one i'm going to just define 2t as u and the integral of this one which is d u is equal to 2 times d t so i'm going to just plug in these values so which is equal to 1 half because d t is equal to d u over 2 i just take out the 1 over 2 outside of the integral ccan to t which is ccan u and d u then so for the integral of this one so we have already known that the integral of cccc and u which is equal to the ln ccccan u and plus tangent u plus c so which is equal to this one so if i just substitute back 2x for you which is the integral 1 over 2 times ln secant 2x not 2x it is 2 t and plus tangent 2 t plus c so i'm going to just plug in these values here so i'm going to get the r of t which is equal to so it is e t plus c1 times i minus cosine t plus c2 times j and plus one half and then ccan 2t plus tangent 2 t plus c3 and times k so so these are this is the integral of the function so in the question also it says when r is equal to zero that means when you plug in, so 4 t is equal to 0, let me just go forward.
04:40
For t is equal to 0, i'm going to plug in 0 for each of the variable.
04:45
That means e to the power is 0 plus c1, which is equal to 2i, so the coefficient of the i is 2.
04:52
That means c1 is equal to e to the power 0 is 1, so c1 is equal to 1.
04:59
And for again, t is equal to 0, the second one, which is negative, 0 plus c2 which gives us the coefficient of j which is 2 in this case so the cosine 0 is again 1 so c2 is equal to 3 and for the next one which is 1 half times and then cken 0 plus tangent 0 plus c3 which give us the coefficient of the k which is 2 so ccc0 is 1 and tension tangent 0 is 0, which is equal to 1 over ln1 plus c3, which is equal to 2.
05:44
So n1 is equal to, so let's find the ln1...