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Lecture

3.7 related rates.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Fri 2/22/08 3.7 - Related Rates So on Wednesday we went over Implicit Di▯erentiation and what it means to implicitly de▯ne a function. Now let’s go on to a bit harder stu▯ and learn about applications of derivatives. This is where the word problems dwell, so I hope you’re ready for them! Today, we shall learn about Related Rates, an application to Implicit Di▯erentiation. And here they are in a nutshell: they’re basically all word problems and each word problem has a few key elements in it: - You are almost always dealing with some geometric ▯gure. It could be a cone, triangle, square, cube, sphere, etc. Once you ▯gure out what this geometric ▯gure is, draw it, label your variables, and list out all the variables from the question. Now think about which formulas you might need that correspond to this ▯gure. Ie, Volume? Surface Area? Area? Perimeter? If it’s a triangle, Pythagorean Theorem? Maybe something with the angle in a triagle (and using the fact that sin(▯) is opp/hyp or something)? Trickier triangle problems might deal with similar triangles (ie, ratios). We’ll do an example like that later. Note: some related rates problems need two forumlas (like maybe both the volume and surface area are needed!), but there are some that might only need one. - How do your given variables relate to the problem (in particular, the formulas you’re using)? You might be dealing with more than one variable! For example, the volume of a sphere is 3▯r . This is already in terms of one variable, so we’re totally ▯ne. We may have other problems, however. If our geometric ▯gure is a cylinder, then we’re dealing with a volume of 2▯rh. Well, there are two \variables" there, but if you think about it for a second, r doesn’t change at all! It remains constant as we ▯ll this cylinder with something. But, if we’re dealing with the volume of a cone, our volume formula is ▯r h. You have two di▯erent variables here! Both the height and 3 radius are changing with respect to time! This is when you would need to pull out a second formula and try to get one variable in terms of another. - Take the derivative with respect to time! These are related rates we’re talking about! Don’t forget that you probably need to do some implicit di▯erentiation! Also, never ever ever plug in any values from the problem until after you have taken the derivative. Just don’t. Make sure after you’ve taken the derivative that somewhere, whatever the question is asking for shows up in your dV new equation! Ie, if it’s asking for the change in volume, make sure that dt shows up somewhere! - Plug in the values and hope that everything works out! Okay, I think it’s time for an example! 1 Example: #16 p. 222. A 17-ft ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of 5 ft/s, how fast will the top of the ladder be moving down the wall when it is 8ft above the ground? Solution: Well, ▯rst thing we should do is draw a picture. Ladder against a wall forms a triangle. Well, here it is!     17  y      x Now let’s list out the variables. The ladder itself is 17 ft long. That ain’t changing. we’re taking the ladder and pulling it away from the wall at a rate of dx = 5. We want to know dy (how fast dt dt it’s moving down the wall) when y = 8. So, what formulas can we use here? Area is useless. Hmm...We have a right triangle, and we could care less about the angle of that ladder...perhaps the Pythagorean Theorem? 2 2 2 17 = x + y Well, both x and y are changing with respect to time, so we actually have two variables we’re dealing with here. So, let’s di▯erentiate! 0 = 2x dx + 2y dy dt dt Now let’s plug in the stu▯! So, dt = 5, y = 8...uh oh! What about that x? How do we get that x? 2 2 2 Pythagorean Theorem! 17 = 8 + x . Therefore, x = 15. Therefore, dy 0 = 2(15)(5) + 2(8) dt dy 16 = ▯150 dt dy ▯75 = dt 8 The answer is negative because the top of the ladder is moving down at a rate of 75/8 ft/s. 2 Example: If a snowball melts so that its surface area decreases at a rate of 2 cm /min, ▯nd the rate at which the radius decreases when it is 5 cm. Solution: Okay, what are our variables and givens? surface area decreases at some rate, so we’ll dA dr call thatdt= ▯2. we want to know the rate the radius is decreasing, so we’re solvidtwhen r = 5. All right. Now, which geometric ▯gure are we dealing with? Well, snowballs are spheres, and it’s talking about the surface area, so that means we need this formula: A = 4▯r 2 Now let’s di▯erentiate with respect to t! dA dr dr = 4▯(2r) = 8▯r dt dt dt And now substitute in the givens dr ▯2 = 8▯(5) dt dr = ▯1 dt 20▯ Therefore the radius of the snowball decreases at a ra20▯of/min Example: Whenever Chuck Norris roundhouse kicks an enemy, his left foot follows along the path of a sem
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