Spooky Halloween Related Rates #1

So here is our spooky ghost. And now I'm going to use this as my Y distance. In its 40 feet, above the ground. Of the distance between Katie and the ghost is increasing, the distance between Katie and the ghost is increasing at a rate of 7 feet per second, how fast is the ghost rising? Okay, well, I can see pretty quickly that this is a Pythagorean relationship. And we're given an X distance of 30, a wide distance of 40, so I'm going to call, I don't really have a variable for this. I'll just call it C and with Pythagorean triples or Pythagorean theorem, I can find that that's going to be a 50 feet. And you're told the distance between Katie and the ghost is increasing. Well, that would be a D.C. DT. And it's increasing at a rate of 7 feet. Per second. Skip into the question, we want to know how fast the ghost is rising, that's your wide distance, so we are looking for DY. DT. Okay, so with any related rates, we want to establish some kind of formula. We have a rot triangle relationship. 

So with Pythagorean theorem, I could say X squared plus Y squared is equal to C squared. I'm going to take the derivative with respect to T so we use implicit for each term. So I'm going to see a D, a two X, DX DT. Plus two Y DY DT equals two C, D, C, DT. Now at this point, we can see what we know numbers four and make some replacements. Keep in mind, we're trying to find DY DT and that is represented in the problem. So plugging in, I have a two, then I'm going to carry my X distance was 30. My DX DT is a negative three. Bring down a plus spring down a two at this moment in time. Y is a 40. DY DT is what we want. Bring it down to two. My C that I calculated at that moment in time was 50, and the given D.C. DT was 7. Okay, fortunately we are down to just one variable so we can do some math. To solve for it. And then work it through to getting DY DT by itself. When you divide, DY DT looks like 11 feet. Per second. Okay, there's problem number one. Okay, number two, Katie's leaving the cemetery. She sees a giant spherical pumpkin balloon. It has a volume of 36 pi, cubic feet. You should have a pie. In the problem there. 

It's increasing at a rate of two feet cubed per second, how fast is the radius of the balloon increasing, and I gave you a volume formula to work with. For a sphere that's four thirds pi R cubed. Well, we can list what we know. And at this moment in time, this balloon has a volume of 36 pi. The volume is changing at a rate of two feet cubed per second. Well, that's the rate of change of volume, with respect to time, or DV DT. I'm going to call it a positive two, since it is increasing. We want to know how fast the radius is changing. So I want to know DR DT. That's our goal. Well, again, it starts with the formula, for a sphere, four thirds, pi, R cubed, and we use calculus to take a derivative with respect to time. That's a DV DT. Remember four thirds pi is a constant. So I'm going to get that out of the way and then take the derivative of R cubed. As a three R squared, DR, DT. We can do you have a choice. You can either do some algebra to try to get this DR DT by itself, or we can plug in what we know. And then hope that we have all numbers. So let's just see what happens here. I know I can replace DV DT with two. Keep my four pi over three, hanging out, let the three hang out eventually that's going to cancel my other three, but I'll do that later. 

My R would be the radius at the time, the volume is 36 pi. That wasn't given to us, but we can find it. Going back to our original volume formula at this moment in time we know we have a volume of 36 pi. But we don't know what that radius is, but I could solve for this. Your pies will divide out. Multiplying by the reciprocal will get R cubed by itself, that happens to be 27. And then the cubed root will be three. So at the moment that the volume is 36 pi, the radius is three. So that becomes my R, which I will square later, and its DR DT that I want. Okay, it's time to clean this up a bit. Combining all these coefficients on the right side, you'll see a 36 pi. DR DT. Dividing, we can call this two over 36 pi. And if you didn't have a calculator, you would stop there. If you are asked to round, then this would be the time where you pick up a calculator and evaluate this, but we're going to say DR DT using a calculator becomes .0 one 8 rounded to three places. And this would be feet per second. And I know it's feet and not feet cubed because this variable here is a radius and radius is one dimensional. 

So I'm going to have a one dimensional unit, and then my second do match up with the time that's in the bottom there. All right, let's move on to problem number three. All right, so the evil Freddy has a giant piece of rectangular taffy. Katie grabs one in, starts pulling. The taffy's going to maintain a rectangular shape, and a constant area of 3600 cm². The length is 90 and is increasing at a rate of two centimeters per second, how fast is the width changing at this moment. Well, let's think about this phrase constant area. The taffy can't change in its area. So, if the length is increasing, but the area must remain the same, then we expect the width to decrease, because as one dimension changes, the other one has to change in the opposite way. So, we're going to expect a negative answer here. By planning ahead, we can have those checkpoints in place to see if our answer is reasonable. So we expect a negative answer to make that area stay the same. Well, we need to start with a formula, and the key to that is the shape of the taffy. So we're talking to area, so I'm going to start with area equals length times width. I would, to get to the right, I do need to take the derivative. Now this one's a little bit different because you've got two variables multiplied, so we have to use the product rule to differentiate length times width. Left side, I'm going to call DA DT. 

Now starting off the product rule, I need derivative of the first DL DT times the second plus derivative of the second or DW DT times the first. Okay, you've got an equation with all letters. Let's see what kind of numbers we could come up with. Rereading. I see that the area of the taffy is a constant 3600. But there is no aerial listed in my equation I have a DA DT instead. I know that the length is 90 at this moment in time, so with a little bit of math, we can divide and find that for the area to be 3600, the width would have to be 40. I know that that length is increasing at a rate of two centimeters per second. Well, that's DL DT, increasing means it's going to be a positive. Two centimeters per second. What we want to know is how fast the width is changing, so that's a DW DT. That's what we want to know. All right, going back to my equation, I need something to replace DA DT with. Well, if you'll recall, the area is constant. What do we know about the derivative of a constant? Got to be zero. 

So I have zero equals, DL DT gets replaced with a two. W would be the width at this moment in time, which we calculated to be 40. DW DT is what we want to know, so that should stay a variable, and links at this moment in time was 90. Okay, so we solve for DW DT by subtracting 80. And then with division, looks like negative 8, ninths. And we're not surprised that it's negative. We expected that. And if we wanted to evaluate that with a calculator, it turned it into a decimal, I would say DW DT is a negative .889. Centimeters per second. I'm using centimeters because this variable W is one dimensional, so it's not centimeter squared just because we're dealing with area. You go with your variables to be consistent. Hey, three down one more to go. Okay, Katie prevails over Freddie and gets home with her bag of loot. Like all kids on Halloween night, she's going to dump it out in the candy forms a conical pile on the floor. The diameter will always be four times the height. The height is increasing at a rate of two inches per second, so we want to know how fast the volume is changing when it's 9 inches high. 

Okay, let's make a list of some things we know. Going back to the beginning or almost the beginning, the diameter of the Powell is always four times the height, that means D is equal to four H, at any given moment in time. I know the height is increasing at a rate of two inches per second. There's your DH DT. The question asks how fast is the volume changing? That means I want to know what DV DT is. And at this moment in time, it's asking about the cone is 9 inches high. So my height is 9. Well, before I jump into a volume formula, I can use the fact that the height is 9. And that the diameter is always four times that, to know that the diameter is going to be 36. Your volume is though in terms of radius. So let's go ahead and list that as 18. Okay, now we can introduce the volume equation. Pi over three times R squared H. We want the DV DT so I'm almost out of time. I have a 15 minute limit, so I'm going to finish this one off here and then start my second video to finish this problem out.