Page mathematics 2450 calculus 3 final exam show all details...

Page Mathematics 2450, Calculus 3, Final Exam Show all details. A correct answer with no work (or unreadable work) counts as zero. 1. Let the velocity vector be $v(t) = \sin t i + e^{2t} j - 2t k$, and the initial position vector be $r(0) = -i + 2j - 2k$. Compute the acceleration vector $a(t)$, and the position vector $r(t)$. Answer:

Transcript

00:01 In this question we have been given the position function of a particle.

00:06 Okay, so it is given as r of t is equals to 3 cosine of t, 2t and sine of t.

00:18 Okay, so with this position function, first we need to find out the velocity function.

00:23 So we know that velocity function is same as r -daz of t.

00:26 So we have to take the derivative, so it becomes minus 3, sine t.

00:31 And here it is one, sorry, 2t derivative will be what? it will be 2 and sine t derivative is cosine of t.

00:40 So this is my velocity function.

00:44 Then we have to find out the acceleration function, okay, a of t.

00:48 So it is the derivative of the velocity function.

00:51 So it becomes minus 3 cosine of t.

00:55 Then i will get zero here and this is minus sine t.

01:00 So here comes out my.

01:02 Acceleration function then we have to find out what is the magnitude of the velocity function so it comes out to be under root of minus 3 sine t whole square plus 2 square plus cosine of t square that is cosas square t so i will get 9 sine square t okay nine sine is and here it is 3 sine t okay so here i will get 3 cost t so it will be minus 3 sine t and here that will get 9 cosquarity.

01:36 So we know that 9 cos and square t plus sine square t, it becomes 9 only.

01:42 So 9 plus 2 square, which will be 4.

01:49 Okay, so 2 square is 4.

01:52 So this comes out to be under root of 13.

01:57 It is the velocity magnitude.

02:00 Then now let's move to the next question.

02:02 In the next question, we have been given the axis.

02:05 Acceleration which is 2i plus 60 j plus 12 t square k and the velocity at time t equals to 0 is given to be i okay and velocity and the position at t is equal to zero is given as 5 j minus 4k correct so from this acceleration we need to find out the velocity correct so how we are going to get that so let us see so velocity will be what we know that it is the integral of the acceleration okay so if we integrate this so this is nothing but component wise if i write two and then we have 60 and here we are having 12 t square okay this is what i am have so let us differentiate it.

03:08 So i will get my velocity as this will be component for me if i write...

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