### Video Transcript

A particle started moving along the π₯-axis. At time π‘ seconds, its position relative to the origin is given by π₯ equals 12 sin of 14π‘ plus nine cos of 14π‘ meters for π‘ is greater than or equal to zero. Find the maximum distance between the particle and the origin π₯ max and determine the velocity of the particle π£ when π‘ equals three π seconds.

Weβre dealing with a particle moving in a single straight line. Itβs moving along the π₯-axis. Its position relative to the origin is also known as its displacement, and this is equal to some expression in π‘. So how do we find the maximum distance between the particle and the velocity of the particle? Well, we can find the location of relative minima and relative maxima of a function by differentiating it and setting it equal to zero. So we can find the maximum displacement or at least the time when the maximum displacement occurs by differentiating π₯ with respect to π‘ and setting that equal to zero.

π₯ is made up of two trigonometric terms, so weβll differentiate term by term. We begin with the term 12 sin 14π‘. And we recall that the derivative of some function sin ππ‘ for real constants π with respect to π‘ is π times cos of ππ‘. This means the derivative of sin of 14π‘ is 14 cos of 14π‘. And so the derivative of 12 sin of 14π‘ with respect to π‘ is 12 times 14 cos of 14π‘.

And what about the derivative of nine cos of 14π‘? Well, for real constants π, the derivative of cos ππ‘ with respect to π‘ is negative π sin of ππ‘. And this means the derivative of cos of 14π‘ is negative 14 sin of 14π‘. So the derivative of nine cos of 14π‘ is nine multiplied by this. dπ₯ by dπ‘ then simplifies to 168 cos of 14π‘ minus 126 sin of 14π‘. To find the location of relative minima and relative maxima, we set this equal to zero, and weβre going to solve for π‘.

Now, at the moment, we have both cosine and sine in the same expression. And so weβre going to recall the identity that links sine and cosine. We know that tan π is sin π over cos π. So letβs rearrange our equation so that we have some expression sin π over cos π, where here π will be equal to 14π‘. Weβre going to add 126 sin of 14π‘ to both sides. Next, we divide through by cos of 14π‘. So our equation is now 126 sin of 14π‘ over cos of 14π‘ equals 168. To make sine over cos the subject, weβll now divide through by 126. So sin of 14π‘ over cos of 14π‘ is 168 over 126. Well, 168 over 126 simplifies to four-thirds, and we can now replace sin of 14π‘ over cos of 14π‘ with tan of 14π‘.

Weβll solve for 14π‘ by finding the inverse or arc tan of both sides. This gives us 14π‘ equals 0.927 and so on. Remember, we have to work in radians. Now, in fact, we know that there are further solutions to this equation. The tangent function is periodic, and it has a period of π radians. So we can add and subtract multiples of π to this value for 14π‘ to find further solutions. We donβt need to though; we will get the same result for each value of π‘. And so instead, we divide both sides of this equation by 14. And we find π‘ is equal to 0.0662 and so on, and this is in seconds.

Now, weβre not going to round our answer for π‘. We found the time at which the particle is at its maximum distance from the origin. And so we need to substitute this into the original equation for π₯ to find the actual value of the maximum distance. And when we do, we get π₯ equals 12 sin of 14 times 0.06 and so on plus nine cos of 14 times 0.06. Now, that actually gives us a really lovely answer of simply 15. Since weβre measuring this in meters, we can say the maximum distance between the particle and the origin, π₯ max, is 15 meters.

Weβre not finished though; weβre asked to find the velocity of the particle π£ when π‘ is equal to three π seconds. And so we recall that to find an expression for the velocity given an expression for π₯, the displacement, we differentiate π₯ with respect to π‘. This is purely because velocity is change in position with respect to time. We can, therefore, go back to some of our earlier working and say that π£ must be equal to 168 cos 14π‘ minus 126 sin of 14π‘. And this means we can find the velocity of the particle when π‘ is equal to three π by substituting three π into this equation. When we do, we get 168 cos of 14 times three π minus 126 sin of 14 times three π. Thatβs simply 168.

π₯ max is 15 meters. And the velocity π£ is 168 meters per second.