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# Suppose you start at the point $(0,0,3)$ and move 5 unitsalong the curve $x=3 \sin t, y=4 t, z=3 \cos t$ in the positive direction. Where are you now?

## $$(3 \sin (1), 4,3 \cos (1))$$

Vectors

Vector Functions

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##### Lily A.

Johns Hopkins University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

all right. In this problem, we have a curve represented by the vector ar 15 with three component functions three sine of T four t and three co sign of tea. And what we want to think about is starting at the 30.3 and moving five units along this curve, and we want to figure out where we are after we've moved five units. So what points do we end up with after moving five units along the curve? So we do want to think about our formula for the length of the curve because we know that this length is going to be five. We move five units. So we're going from from our starting point to our ending point, we'll figure out what that interval is. In a moment. We still need to find the magnitude off the derivative of that vector and then integrate that with respect to t. So let's go ahead and find the magnitude of the derivative first, um, So first we need to take the derivative with respect to T, which is going to give us three co sign of tea four and negative three sine of t and then we want to take the magnitude of that. So we're going to get magnitude of our prime of tea is equal to the square root of our first component function squared. So we get nine co sine squared of tea plus 16 plus nine sine squared a t going Teoh Rearrange that a little bit. So we're gonna end up with 16 plus nine times and factoring the nine out of the first and last term. So nine times co sine squared of tea plus sine squared of tea which we know from our trig identities is equal Teoh one. So we end up with the square root of 16 plus nine, which is the square root of 25 or five. We also want to think about, um, the the starting point for T s 01 of the ways that I would do this is look for the simplest of our three component functions. So I would say that our simplest is 40 and that is our our why component. So if I look in the starting point, why is equal to zero? So I could think about the fact that when why is equal to zero for key is equal to zero. So that means that are starting value for T is going to be zero If we saw that for team. So we want to find out. Um what the That's our starting point for tea. We want to find out what the ending point is for tea, so we can ultimately calculate, um, are triple coordinate triple. So now that we know what our starting point is, we want to figure out we'll set up our expression for the length of the curve. So we would be going from zero to t because we don't know where we're ending at five units. The derivative are the magnitude of the derivative of vector Are we found was five. So this is going to be five or enter it great with respect to you, which is just our dummy variable, And then we can, right, this is five you and use the fundamental theorem of calculus to evaluate this. So we have five times t times there minus five times zero, which is just zero. So we know this much and we know that we have moved five units along the curve. So in this case, the length of the curve that we're interested in is five. And then if we take that information, this right here tells us that we end. Where are parameter? T is equal to one. So if we want to find out the coordinate triple that we end up with, we can plug that into your three component functions. So are three component functions are three sine of T. It's four of tea and three co sign of tea. So then we can look at our ending point, which is t equals one and value each of those. So we end up with three times the sign of one. Why we just end up with four times one or four frizzy. We end up with three co sign of one. So if you want to write that as a coordinate triple, we can say that we end up at the point. Three sign of one for three co sign off one And again, that is, after traveling five units along the curve represented by our of team

Campbell University

#### Topics

Vectors

Vector Functions

##### Lily A.

Johns Hopkins University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp