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Consider the following position functions.

a. Find the velocity and speed of the object.

b. Find the acceleration of the object.

$$\mathbf{r}(t)=\langle 13 \cos 2 t, 12 \sin 2 t, 5 \sin 2 t\rangle, \text { for } 0 \leq t \leq \pi$$

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Johns Hopkins University

Harvey Mudd College

Baylor University

University of Nottingham

Hi. We're given the position Vector are as a function of tea, with each of his components being a triggered a metric function. And we're acts to determine the velocity vector, which is simply the time derivative of the position vector you could simply but it is a d r t d t. But we have a problem because the trigger metric functions are actually in terms of to t and not t so it's not co sign t is co signed up to t, so we would have to then use the chain rule to go and get the derivative for the position Vector and the Chamber of Simply States that if you are taking the derivative of a function of a function with respect to DT, you would have to break it out and do the derivative of that function relative to the function and then take the function itself and then do that every bit of it. Well respected D t. And in our case, the G two, it's just simply to t Okay, so if we take the derivative of each component of the position vector, it would look like this where we would take the 13 co sign to tea with respect to T, and that would just be the diverted about The cold sign is a negative sign, so that's 13 sign up to t times that derivative up. The GMT, then is the derivative to tea with respect to T, and that would just simply be, too. And then same would follow for the other components. The X then would be the derivative of the sign of to tee times 12. So that would be 12 times to co sign of two t Times two and why Z component would be five times co sign of two t I'm still again. So who were to take all those components and put them in the positional vector for of the value that perform? It would be a negative 26. I'm signed of two T 24 times to co sign of two T and then 10 times to co side of two t so that let's go back and clean that up with. That's a little bit more ledge. It'll to t. So that would be your velocity vector that you where acts to determine the next question. That's part of this problem is to determine the speed. So we have the velocity vector here, and the speed then, is going to be the magnitude of this velocity factor that you will recall. The magnitude of a vector is equal to psalm of the squares of each component, and then you take this square root. So basically, I've written that ahead of time already, and the magnitude of the velocity vector then would be the square root of 26 squared signs Square two T 24 square coastline square to T plus 10 squared Co. Side of to T and all of that you would take after you've computed that you would take the square root. I'll let you go through and do the math, but combining these last two terms you would get 676 sides square to to T plus 676 coast that square to t. You can factor out that 6 76 and you would end up with 6 76 times the signs square to to T plus coast. That square to to tea, which you should recognize is being the radius for the unit circle, and that's one. So then you would end up with the speed being the square root of just simply 676. And if you punch that on your calculator, you'll see that the speed or the magnitude of the velocity vector he's 26. So that's the answer. For the second question that we were asked. The last question we asked was to determine the acceleration vector for the object and the acceleration again, just similar to, uh, how we got the position vector, how we got the velocity vector waas to go and do the derivative so well. But the 26 back here again. So the acceleration vector then, is again the time derivative of the velocity that here, and we'll go through the same process that we went through to determine the velocity vector as far as determined the directive where we would take the derivative of the trigger metric function times the derivative or the function of the sine function so intense in this case, then the acceleration vector is a minus 52 coz i to t 48 times a sign of two t and then the Z component, then is 20 times a sign A to T, and that is what your velocity vector your acceleration factor would be. Circle that and

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