| [0.91s -> 13.94s] Number 1. A 17 foot ladder is leaning against a building. The foot of the ladder is 8 feet from the base of the building and it's sliding away from the building at 3 feet per second. |
| [14.13s -> 28.75s] How fast is the top of the ladder sliding down the wall of the building? Well, let's find out. So let's start with a picture. So let's say that's the building. This is the ground. And let's say this is the ladder. |
| [30.19s -> 40.43s] Now let's call this X Y and let's say the length of the ladder is Z Now let's write down what we know |
| [41.55s -> 53.78s] So we have a 17 foot ladder, therefore z is 17. The foot of the ladder is 8 feet from the base, so that's the value of x. |
| [54.77s -> 58.61s] And we don't know the value of y. |
| [59.50s -> 72.53s] Now, we know that the ladder is sliding away from the building, that is, it's sliding in this direction, at 3 feet per second. So every second, x is going to change by 3 feet. |
| [73.10s -> 83.76s] So that's dx dt. That's the rate at which x is changing. So let's write that here. So dx dt is 3 feet per second. |
| [86.35s -> 96.98s] Now we don't have the value for dy dt But in part a that's what we're looking for how fast is the top of the ladder sliding down the wall? |
| [97.55s -> 111.34s] So as the ladder slides to the right, it's also sliding down the wall. And so y is changing as well. Notice that y is decreasing while x is increasing. Because x is increasing, |
| [111.34s -> 124.02s] dx dt is going to be positive. And because y is decreasing, we should expect a negative answer for dy dt. So let's go ahead and calculate dy dt. |
| [124.75s -> 133.55s] So we need an equation that relates x, y, and z. So notice that we have a right triangle. |
| [134.03s -> 147.98s] And according to the Pythagorean theorem, c squared is equal to a squared plus b squared. Now c is the hypotenuse, so c matches with z. And we can say a is x, b is y. |
| [148.27s -> 162.48s] So z squared is equal to x squared plus y squared. Now, before we could differentiate both sides with respect to time, we need to calculate the value of y. So z is 17. |
| [163.28s -> 177.36s] x is 8, what is y? 17 squared, that's 289. 8 times 8 is 64. 289 minus 64 is 225. |
| [178.19s -> 192.94s] And now we need to take the square root of both sides. The square root of 225 is 15. So now that we have the value of y, |
| [193.33s -> 204.62s] we should differentiate this equation with respect to time. So the derivative of z squared is what? Is it 2z times dz dt? |
| [204.94s -> 216.88s] Now, notice that Z is not a variable but a constant. Z cannot change. So therefore, dz dt is 0. So anytime you find the derivative of a constant, it's going to be 0. |
| [217.23s -> 227.28s] So this is going to be 0. It's equal to 2x times dx dt and the derivative of y squared will be 2y times dy dt. |
| [228.24s -> 233.49s] So we could divide everything by 2 0 divided by 2 is 0 so we can get rid of this |
| [235.38s -> 244.11s] x is 8, dx dt is 3, y is 15, and we need to calculate dy dt. |
| [244.72s -> 257.50s] Now, 8 times 3 is 24, so let's move that to this side. On the left side, it's going to be negative 24, and that's equal to 15 times dy dt. So if we divide both sides by 15... |
| [257.50s -> 268.66s] we can see that dy dt is negative 24 over 15. But we can reduce that. So let me just reduce it somewhere on the left side. |
| [269.68s -> 282.29s] 24 is 8 times 3. 15 is 5 times 3. So we could cancel with 3. So therefore, dy dt is negative 8 over 5. |
| [283.50s -> 285.97s] And that's it for part A. |
| [291.18s -> 299.09s] Now, let's move on to Part B. How fast is the area formed by the ladder changing at this instant? |
| [301.62s -> 314.03s] So what is the area of a right triangle? The area of a right triangle is 1 half base times height. In this problem, we can see that the base is x and the height is y. |
| [314.74s -> 323.57s] Now we have dx dt and dy dt. So in this form, we can go ahead and differentiate this equation with respect to time. |
| [325.49s -> 338.45s] The derivative of A is going to be dA dt, which is what we're looking for. And then we need to use the product rule. So let's say if we have F times G. |
| [339.34s -> 346.26s] It's going to be the derivative of the first part times the second plus the first part times the derivative of the second. |
| [347.28s -> 355.02s] So let's say the first part, f, we're going to say it's 1 half x. And the second part, g, is y. |
| [356.08s -> 369.42s] So f prime, the derivative of 1 half times x, is going to be 1 half times the derivative of x, according to the constant multiple rule. The derivative of x is 1, but times dx dt. |
| [369.65s -> 383.34s] And then we need to multiply by the second part, which is g, and g is y. And then plus the first part, which is not going to change, so that's 1 half x, times the derivative of the second part. |
| [383.89s -> 389.46s] which is g prime, the derivative of y is 1, times dy 18. |
| [396.21s -> 405.58s] Now all we need to do is plug in the information that we have. So this is going to be 1 half times dx dt, which is 3. |
| [405.97s -> 418.64s] And then dy dt, I mean not dy dt, but y, that's 15. And then plus 1 half times x, which is 8. And dy dt is negative 8 over 5. |
| [418.70s -> 428.98s] And I forgot to write the units for dy dt, so let's talk about that. So what's the units for y? x and y have the units feet. |
| [429.26s -> 437.84s] And then the unit for time is going to be seconds. So dy dt is feet per second. |
| [439.44s -> 447.76s] Now let's go back to this problem. So we have 3 times 15, which is 45, and half of that, that's going to be 45 over 2. |
| [449.33s -> 458.45s] Half of 8 is 4, and 4 times negative 8, that's going to be negative 32. So this is negative 32 over 5. |
| [458.80s -> 464.46s] So we need to combine these two fractions, and so we have to get common denominators. |
| [465.42s -> 475.92s] The common denominator between 2 and 5 is going to be 10. So we need to multiply the fraction on the right by 2 over 2 and the one on the left by 5 over 5. |
| [476.53s -> 489.58s] So what's 45 times 5? 40 times 5 is 200. 5 times 5 is 25. So this is going to be 225 over 10. 32 times 2 is 64. |
| [490.86s -> 496.78s] And if we subtract 225 by 64, that will give us 161. |
| [497.17s -> 511.25s] So, DA over DT is going to be 161 over 10, and the units is going to be square feet, because we're dealing with area, per second. So, this is the answer. |
| [512.53s -> 513.97s] for Part B |
| [524.78s -> 536.88s] Now let's move on to Part C. Find the rate at which the angle between the ladder and the ground is changing at this instant. So where is the angle between the ladder and the ground? |
| [537.42s -> 545.36s] So here is the ladder. Here's the ground. Therefore, the angle between that is right here, which we'll call theta. |
| [546.22s -> 559.31s] Now, we need to relate theta to x, y, or z, and we can use any of the three trig functions, sine, cosine, or tangent. However, one of them I would not recommend using. |
| [559.66s -> 565.52s] since it involves a lot of unnecessary work. Now, let's review Sokotoa. |
| [565.97s -> 579.66s] Hopefully, you remember the principles taught in trig. So let's start with so. This tells us that sine theta is equal to the opposite side divided by the hypotenuse. Opposite to theta is y. |
| [579.66s -> 591.25s] And the hypotenuse, which is across the box, that's Z. That's the longest side. So sine theta is Y divided by Z. Now, Ka tells us that cosine theta... |
| [591.73s -> 602.61s] is equal to the adjacent side, adjacent to theta is right next to it, that's x, divided by the hypotenuse, z. And toa, tangent theta, |
| [603.15s -> 617.39s] is equal to the opposite side, y, divided by the adjacent side, x. Now, keep this in mind. x and y are variables because they're changing. z is a constant. |
| [618.74s -> 632.80s] Now, for sine, you have a variable and a constant. You could simply use the power rule to differentiate that. And the same is true for cosine. You have a variable and a constant. For tangent, you have two variables. And so you need to use the quotient rule. |
| [632.80s -> 640.66s] if you're going to use tangent theta. And let's avoid the quotient rule, so let's not use tangent theta. I'm going to use sine theta. |
| [650.90s -> 665.33s] So sine theta is equal to the opposite side of theta divided by the hypotenuse. So I'm going to rewrite y over z as 1 over z times y. |
| [666.58s -> 674.26s] So you can see what we have is a constant times a variable. Now let's differentiate both sides with respect to time. |
| [674.90s -> 688.24s] So the derivative of sine theta is cosine theta. And then, according to the chain rule, we need to differentiate the inside function theta. The derivative of theta is 1 times d theta over dt. |
| [689.62s -> 701.01s] Now let's use the constant multiple rule. So the derivative of a constant times y is going to be the constant times the derivative of y, which is 1 times dy dt. |
| [702.77s -> 704.82s] Now, what is cosine theta? |
| [705.36s -> 719.06s] If you recall, we said that cosine theta is equal to the adjacent side, which is x, over the hypotenuse z. And we have x. x is 8 and z is 17. So therefore, cosine theta... |
| [719.06s -> 729.30s] is 8 over 17. Our goal is to calculate the rate at which the angle is changing. So we need to calculate d theta dt. |
| [729.71s -> 742.00s] Z, we have that, is 17. And dy dt is negative 8 over 5. So now let's multiply both sides. |
| [746.19s -> 748.21s] 17 over 8 |
| [751.02s -> 761.42s] If we do so, notice that the 8s on the left cancel and on the right will cancel. And at the same time, the 17s will cancel on both sides. |
| [763.28s -> 775.76s] So just by doing that simple move, we have the answer. So the only thing that's left over is d theta dt. And as we can see, it's equal to negative 1. |
| [779.18s -> 793.26s] over 5. And that's it. So that's the answer for part c. d theta dt is negative 1 over 5. And let's talk about the units. The unit for d theta dt |
| [793.90s -> 800.91s] So the angle is in radians, and the time is in seconds. |
| [802.93s -> 809.74s] And that concludes this lesson. So that's it for this video. Thanks for watching. |
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