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# How fast is the angle between the ladder and the ground changing in Example 2 when the bottom of the ladder is 6 ft from the wall?

## The angle between the ground and the ladder is decreasing at the rate of $-\frac{1}{8}$ Radians/ sec

Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

All right, We've got the question here that describes a ladder that is being inclined onto a wall. Okay, this line here is the ladder. This here is the wall, and we have an angle data that shows the relationship between the two. And if we're if we're told that the bottom of the ladder 6 ft from the wall. Hey, and we also know that the latter is given to us. If we look at example two, the latter is 10 ft long. Then we want to determine how fast the angle theater is. Data is changing. Aziz Felax er is pulled back. Okay, So ladders pulled back from the bottom. How fast is that angle changing? Excuse me. All right. So, first, what we can do is we can use the potassium theory, Um, to get this side here of the wall So we have X squared plus y squared equals C squared. And here we know our Z squared is the high pot. News 10 squared is equal to 100. And then our X squared is six here. So six squared is 36. Now we saw for Weiss. Squared was to 100 minus 36. And why will be swear of 64? Because we know 100 minus 36 is 64. So we're left with eight. Beat Yes, or a fee will be our well Why? All right now waken use a sine function to describe data once they will sign of data is equal to the opposite. Which is why here, over hi pot news, which is 10. And we know what our why is we just calculated for it. So we'll say That's 8/10. And so we take these sign Excuse me Way, differentiate! Now we can calculate for the change in data, so we would dio um Actually, it will probably be easier if we write out the cosine function as well, because we know whenever we derive a sign, we end up getting a coastline. So So let's also calculate for cosign data, so cosign data would be adjacent overhype Odd news. Just six old time. Okay, now let's go and differentiate that co sign so we can get a negative sign Data data over DT once again just to clarify, we are taking the derivative of that so that we could get the change of data over the change in time. And we would have all of that equal to a Well, let's so the variable that is changing right is this distance here? I just I've just realized that so since that distance is changing, we can also keep it as a variable X so that when we do take the derivative way will get X over data. Excuse me. X over 10 and then you take the derivative ex. Excuse me, That'll go toe one and then you have DX over. DT stylist is acting up. There it goes. We have t X over. What is this? Some reason my stylist is acting up, so I'm gonna go and move it over to this side. For some reason, it doesn't like that spot there. So were saying that this is equal to this. Here. It should be 1/10. The X over teaching. Okay, now, if we go and move everything over on on Lee, leave the data over duty on one side, Then we'd have a one over in negative sign data multiplied by 10. Oh, okay. And then that would be multiplied by DX DT Oh, sorry. For one some reason is it's not liking that middle, this middle area here, so it won't let me draw there, so I'm gonna move it over. So we have negative one over sign. Excuse me. 10 side, they don't. 50 X. You too. All right. Now we're told that the the change of X as it's moving across the ground, it occurs at one feet per second, and we know sign data we've already calculated for over here. Right? Gordy calculated that if 8/10 so we can plug in all the values now we have You have d date over. DT is equal to negative 10 times, 8/10. These cancel out well supplied by one because we're told that it's when we got 1 ft per second Sort of changing data. Finally, over time, we'll be able to negative one over eight radiance per second. All right? And that will be our final answer there. All right, well, I hope that clarifies the question. Thank you so much for watching

The University of Texas at Arlington

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