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Aim: How can we apply what we know derivatives to physical concepts? By: Nihir Shah.

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Presentation on theme: "Aim: How can we apply what we know derivatives to physical concepts? By: Nihir Shah."— Presentation transcript:

1 Aim: How can we apply what we know derivatives to physical concepts? By: Nihir Shah

2 Physics Concepts: Textbooks and Wikipedia will give you tough definitions. Let me give you simple definitions. Velocity: how fast a thing is moving Accelaration: change of velocity as time passes by. Like sometimes, you go speed up when driving, sometimes, you get tired, and slow down. Distance- how much the object travels.

3 How do we solve Velocity problems: A racecar goes by the following formula: S(x)=12t-6. (t is in seconds) The distance in meters. a.Find the velocity function Its just the first derivative, relax V(x)=12t. b. Find the velocity, when t=7 Easy??? No, well time travel back to Math A!!! Its as simple as plugging in, because we simply want to find the speed during a specific seconds. So, it follows: V(7)= 12(7)= 84 meters/ seconds. (Put your units, or else Mrs. Delacruz will get angry with you )

4 Acceleration problems: Now, dont be scared just because it sounds like a word you havent understood! Memorize this: Accelaration second derivative. Say it to yourself, accelaration second derivative Comeon!!! Example: The Exxon Mobil Corporation moves its enormous oil tank represented by the following S(x)=12x 2 + 6x+3 Try it, just remember accelaration, second derivative

5 Answer: Did you get it??? Well, lets find the first derivative first. We get V(x)= 24x+6. Now, what did I tell you, the accelaration function has? The Second derivative Yes, good job! A(x)= 24. Dont worry, if you think you got it wrong. You didnt. Some acclerations are constant, and for linear functions, they sometimes dont exist.

6 Distance problems Distance- is the third type of problem that ETS likes. Heres the deal: You take the first derivative (notice that we always do so), set it equal to 0. We get these points called the critical points. Then, well use these points as well as the endpoints and plug em in. Then will do a strange way of subtraction, which Ill show you later. Lets take an example. Alex wants to determine the total distance traveled during the following interval. S(x)= 12t 2 -10t+3 [0, 5] the distance in miles V(x)= 24t-10. Set it equal to 0. Your e going to get t= 2.4s. Now, its time to substitute. Remember its the interval numbers and the critical point. Item1. S(0)= 3 miles Item 2. S(2.4)= 12(2.4) 2 - 10(2.4)+3= 48.12 miles. Item3. S(5)= 253 miles. Now, its the weird subtraction method. So… Subtract Item 1 from Item 2. So its 48.12 miles-3 miles or 45.12 miles. Next, we do Subtract Item 2 from Item 3. So, its 253 miles-48.12 miles or, 204.88 miles. Now, take the answers that we got, and 45.12 miles+204.88 miles or exactly 250 miles. Amazingly, I made this problem up!!!!!!

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