Vectors

Vector Functions

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

all right. So we want to do is evaluate the length of the curve as described by vector function R T, which we can write in terms of its three component functions. Teas, three co sign of tea and three sine of t. Let's defined on the interval from negative five, 25 So to do this, we want to be able to use our formula for the length of occur as described by our of t. So this is L, um, is equal to the integral from A to B of the magnitude, the derivative of the vector function with respect to t. So the first thing we want to do is take the derivative of our vector function. So we have our prime of tea and the way we do this is just taking by taking the derivative with respect to t of each of our component functions. So we'll have one for the first negative three sign of tea for the second and three co sign of tea for the third. And then if we take the magnitude, uh, that derivative is it's going to be the square root. Uh, are first component functions squared, plus the second component functions squared, plus the third component function squared. And if we simplify that a bit further, we get one plus positive nine sine squared of tea plus nine co sign scored of tea. And ultimately, we want to be able to integrate this function, so it's in our best interest to get a simplified as possible. So what we're gonna do is we're gonna factor out of nine. From the 2nd 2 terms, we're gonna end up with one plus nine times sine squared of tea, plus the plus co sine squared of tea. And then if we use our knowledge of triggering a metric identities, we should be able to see that sine squared t plus co sine squared teeth. That's one of our category and identities, and that equals one. So what we end up with is squared of one plus nine or squared of 10 and then finally, we want to evaluate the length of the curve using the formula in the top right of the screen. So this is gonna be l equals the integral from negative 5 to 5 of the magnitude of our derivative, which is squared of 10 with respect to t so this ends up being square of 10 times T evaluated from negative 5 to 5 and applying the fundamental theorem of calculus. We have five times square root of 10 minus negative, five times square root of 10 which gives us 10 times the square root of 10. So the length of the curve was described by the vector function. R T is 10 times the square of 10.

Campbell University

Vectors

Vector Functions

Lectures

Join Bootcamp