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# According to the model we used to solve Example 2, what happens as the top of the ladder approaches the ground? is the model appropriate for small values of $y?$

## $\frac{d y}{d t} \rightarrow-\infty$ which is practically impossible.

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Mm, yeah. According to the model used to solve example, two were asked what happens as the top of the ladder approaches the ground And is the model appropriate for small values of why so an example to we were given that the latter of 10 ft long rests against a vertical wall. Bottom of the ladder was sliding away from the wall at a rate of four ft per second. Slayers off. And we found how fast the top of the ladder was sliding down the wall when the bottom of the ladder was six ft from the wall. And so if you draw a figure but something like this, we have the height along the wall of the ladder, right? Which is why we have the length away from the Wall X. And then we have the length of the ladder, which is just 10. No applying Pythagoras identity This triangle Well, we found that X squared plus y squared equals 10 squared. A differentiating this equation with respect to the T. Uh huh. Yes, we have two x dx DT plus two y yes, dy DT equals zero. And so solving for dy DT dy DT is equal to negative X over y so times dx, DT, some like all yes, give me sexy. Now if the top touches the ground this represents why approaches zero. And so we see that dy DT will approach mhm. Who's the the rain? Well as why approaches zero. This approach is negative. Infinity as why approaches zero. You see, I can't see anything Honestly. Therefore the rate at which mhm the top of the latter is falling is infinity. Of course, also I want to be So is it right? Yes. 16 her facing you saw for Okay, The x c t We have d y t t times y equals negative x e x t t. Yeah, it's such yes, let's just stare at. And from this it follows that No, do it. Yeah, like this Where you for you, We It's just That's class. Well, since dy DT approaches pretty so So So I tried to but then I was like, mhm like other news. Let's see. Yeah, it says we have x dx DT approaches zero as y approaches zero and therefore the x e t must approach zero is why approaches zero. Did you say so? we have both of these limits to consider that. Thanks, Isis. So we have the rate at which, in fact, and I think we should the ladder is moving away from the wall ST Sure is. Zero means that it's not moving away from the wall. Of course, both of these results are impossible. Refused? Like Russians at the end. Yeah. Uh huh. Yeah. It was like, therefore, it follows that the model is not applicable. Two for small Y yes, but, I mean

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