WEBVTT
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Okay, folks. So now we're gonna take a
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look at this problem. Problem number 17 on your
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book. Um, this is a line growth.
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We have this function app, which is a function
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of position. Let me write down the function here
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. Um X plus y plus z divided by X
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squared plus y squared plus eggs eastward. Okay,
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so we had this function here, and we would
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like to integrate over, integrate over the path which
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is given by our parameter prized by the variable t
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on. I'm gonna write it this way. We
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have tea for X and T for J and T
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for for K X, Y and Z are both
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functions of tea, and they're both the same function
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. They're both t um and we're gonna t is
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going to range from from a all the way to
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be. Um, so this is a very simple
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problem. Um, we're just going to write it
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this way. We're gonna you integrate along the curve
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, uh, of f multiplied by DS. That's
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the That's the earth. That's the idea. Um
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, along the curve, because F is a function
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of x, y and Z, but X,
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y and Z here all functions of tea. So
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let me rewrite the function as a function of teeth
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So X as a function of tea is just tea
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. Same as why Samos Z so f can be
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written in a very simple way. So for the
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nominator we have t squared plus t squared plus t
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spurred. Okay, Multiplied by DS if you remember
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what DS is, the s is the infinite asthma
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line segment which I could write with respect to t
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as the square root of the three. I guess
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you can call it velocities. Some people you can
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you can kind of visualize as a velocity because it's
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just, um the exit e plus do I d
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t pleasant Easy, DT. But you know,
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it's just rid of change of X, y and
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Z with respect to the variable t. So you
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can kind of visualize as a philosophy. That's how
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I visualize it. X is a function of desires
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TV but so DT So the derivative of tea with
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your subjectivity is just one. So one squared is
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still one same as why, same as easy.
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They're all the same function and you take the derivative
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of the function with respect to you. Get one
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. All right. So, uh, and don't
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forget the integration limits, which is between a and
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be That's ah, that's really the basic idea we
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have done. The way we have done, like
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, 80% of the problem there are. All we
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have to do now is to simplify it and see
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what we get. So we have square root of
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three, which I'm gonna pull out of the integral
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because that's a constant. That's a constant. Um
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, we have for the denominator, we have three
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d. I mean, for the new murder,
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we have three t over three t squared DT.
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Okay, so, you know, we're gonna have
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ah, route three one over t b t.
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Which I'm going to rewrite as t of to the
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power of to the power of negative one. You
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feed. You didn't grow on that. You get
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natural log. Right? Um, natural log of
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tea. Um, from a to be No,
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we're done with this video. Basically natural. Log
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of ah, be over a notice here. I
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I kind of jumped the stuff here at the step
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it I could have done Is this square root of
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three of natural August B minus natural log of a
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. But this I just simplified it into this.
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They're the same thing. It's one of the properties
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of logarithms, All right, I think that's it
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for this video. This right here is our finally
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answer for this problem from number 17. And we're
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done. No, this year that the A is
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is actually strictly bigger than zero. That's that's why
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they they explicitly told you that a has to be
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bigger than zero. Because if it's not, we're
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running into a problem here because we have a on
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the denominator here and we cannot divide by zero.
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So a must be strictly bigger than zero. All
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right, that's it for this video. Thank you
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very much in boy.