Slideshow Transcript

1. Slide 1: By Suzanne Gomes
2. Slide 2: Invasion! Related Rates
3. Slide 3: Question • A surveillance hot air balloon is drifting along when it’s cameras detect a strange object in the distance. The worker at the control tower realizes that the object is a spaceship! However, before he can get a closer look, the equipment mysteriously fails… the man decides to use the information transmitted in the instant before the breakdown to calculate the speed of the UFO, so he can send someone to investigate. The balloon was 2km below and 6km to the West of the UFO. The balloon was moving upward at a rate of 144 km/h, and the distance between the two objects was increasing at a rate of 2016/√10 km/h. How fast was the UFO travelling in km/h?
4. Slide 4: x = 6km dx/dt = ? y= 2km dy/dt = -144km/h z = 2 √10 km dz/dt = 2016/ √10 km/h Negative, because it’s going towards the 1. To solve this problem, 2. You can then use UFO, thus decreasing you must first find an this to find the the equation to relate x, y, and distance z. distance. z. x² + y² = z² 6² + 2² = z² z = 2 √10 km
5. Slide 5: 3. The next step is to get an x² + y² = z² equation that relates the rates of change of the three distances. 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt) You can do this by implicit differentiation, and by using the chain x(dx/dt) + y(dy/dt) = z(dz/dt) rule: F’[f(g(x)] = f’[g(x)]*g’(x) (If you consider x² to be a composite of functions, then the derivative of the inner function ‘x’ in relation to time is dx/dt, and the same is true for (6)(dx/dt)+(2)(-144) = (2 √10 )(2016/ √10) y and z.) Now you have an 6(dx/dt) = 4320 expression relating all the velocities, and dx/dt = 720 km/h all you have to do is plug in the values given in the Answer: The UFO is travelling East problem to solve for the at a
6. Slide 6: Giant Rectangle of Doom! Optimization Problem
7. Slide 7: Question • What is the maximum possible area of a rectangle, with sides parallel to the coordinate axis, bound by the graphs of the following functions (to four decimal places): a(x) = -(1/2)x² + 2 b(x) = 2x – 4 c(x) = -2x – 4
8. Slide 8: 1. The first step is to draw a diagram of the problem. The rectangle whose area you want to maximize is shown: 2. To maximize the area, we must first find an equation describing the area. case, the rectangle is In this symmetrical on either side of the y-axis. If we call the distance from x = 0 to one of the vertices ‘x’, the 3. The area of the rectangle can be written entire base can be written as ‘2x’. as: A(x) = base * height The height ‘y’ of the rectangle is = 2x[(-1/2x²+2)-(2x-4)] the distance from the top =2x(-1/2x²-2x+6) function to the bottom function. Written in terms of x, it is a(x)- A(x) = -x³-4x²+12x b(x).
9. Slide 9: Solution A We are looking for the A(x) = -x³- maximum area. One way to 4x²+12x do this is using the derivative of A(x). A’(x) = -3x²- x = -b ± √(b²-4ac) 8x+12 Use the quadratic 2a formula to find the values of x where A’(x)=0. The x = -(-8) ± √((-8)²-4(-3)(12)) maximum value A(x) will 2(-3) be found at one of these x = 8 ± √(64+144) critical numbers. -6 x = 8 ± 4 √13 -6 x = 4 + 2 √13 x= 4-2 √13
10. Slide 10: Solution A x = 4 + 2 √13 x= 4-2 Now it is important to remember √13 the context of the problem. We 3 are looking for an area, so the 3 value of x (which describes the length of the sides) can’t be ≈ -3.7370 ≈ negative. 1.0704 Just to confirm that the remaining zero is a maximum, wemaximum on the parent function do a line test. A is denoted by a zero at some value ‘x=a’ on the derivative, where There is a maximum on the when x<a, f’(x) >0, and when x>a, parent function at x = 4 - 2 f’(x)<0. √13 3
11. Slide 11: Solution A We are answering the question x= 4 - 2 √13 ‘what is the greatest possible 3 area of the rectangle?’ What we have is the value of x at which this maximum is found. A(x) = -x³-4x²+12x To find the final answer, substitute that x-value back into = -( 4 - 2 √13 ) ³ -4( 4 - 2 √13 ) ²+12 ( 4 - the equation that describes the 2 √13 ) 3 3 area of the rectangle. 3 = 7.0354u² The maximum possible area of the rectangle is 7.0354u².
12. Slide 12: Solution B Since we are not looking for the exact value of the maximum area, there is a much quicker way of finding the answer. 3.The initial steps are the same: make a graph and use it to find an equation A(x) that describes the area of the rectangle. A(x) = -x³-4x²+12x 6.Now you can simply use your calculator to find the maximum value. Graph the equation A(x), and press 2nd Calc, 4:maximum. The y-value you get is the maximum value. This is also a good way to check that you did the problem correctly. A(x) = 7.0354u²
13. Slide 13: Revolution Evolution Volumes around x-axis
14. Slide 14: Question • What is the volume of a solid created when the area bound by the graphs of f(x)=x²+2 and g(x)=x+4 is revolved around the x-axis, from x=0 to x=3.
15. Slide 15: When the graphs are rotated around the x-axis, the cross-sections of the solid will form cylinders with radius equal to the value of the upper graph. The cylinders will contain holes with radius equal to the value of the lower graph. For this reason, the area has to be found on two different intervals. The length of the cross sections in this case (the change in x) are decreased until they approach zero. So they essentially form an infinite number of washer shapes like the ones at right. If you add up the areas of all the washers, you’ll know the volume of the solid.
16. Slide 16: 1. Find out where the two f(x)=x²+2 g(x)=x+4 functions intersect, to determine the intervals on which you must x²+2 = x+4 integrate. In this case, as you x²-x-2=0 are only looking for the area x = -1 , 2 from [0,2], the intersection at x=2 is the only one that’s necessary. V₁ = ² πr² - πr² Now you need an equation ₀ describing the area of the washers. =π₀ ² (x+4)² - (x²+2)² The area of a circle is πr². The area of the washers is the total area of the circle (with radius equal to the upper function) minus the area of V ₂ = π ³ (x²+2)² - (x+4)² ₂ the hole (radius equal to the lower function). By substituting the appropriate functions in for r, you get two equations, on the intervals [0,2] and [2,3].
17. Slide 17: Now solve the integrals: V= π (x+4)² - (x²+2)² V= π ³ ₀² (x²+2)² - (x+4)² ₂ = π₀ ² -x⁴-3x²+8x+12 = π₂ ³ x⁴+3x²-8x-12 = π[-x⁵/5-x³+4x²+12x]² ₀ = π[x⁵/5+x³-4x²-12x]³ ₂ = π(128/5 - 0) = π[18/5 – (-128/5)] =128π/5u³ =146π/5u³ The total volume of the solid = 128π/5u³ + 146π/5u³ is the sum of the two volumes. = 274π/5u³ The volume of the solid is 274π/5u³.
18. Slide 18: Solution b 1. Your calculator can be used to find the 1. V= π ₀ ² (x+4)² - (x²+2)² volume of the solid as well. Recall that the volume of the first segment of the solid is expressed by the equation shown. 2. (x+4)² - (x²+2)² 2. Plug the equation describing the area of the washers into Y1. 3. 25.6 3. Press 2nd Calc, 7 to find the integral. Enter the appropriate interval [0,2]. The calculator will shade in that area on the graph, and give you the value 25.6. 4. 25.6 π u³ 4. Multiply that value by π, as it was not included in the integral. This is the volume of the yellow segment of the solid. 5. 29.2 π u³ 5. Repeat steps 1-4 using the equation for the volume of the green segment. 6. 54.8 π u³
19. Slide 19: Dimension Hopping Differential Equations
20. Slide 20: Question • A magical imp from an alternate dimension is moving through various planes of existence (or sometimes non- existence, alternate universes can get pretty strange). His acceleration is described by the equation dy/dx = cos²x · sinx · y Find an equation that describes the imp’s velocity if at time zero he is moving at 2e⅓ m/s.
21. Slide 21: 1. dy/dx = cos²x · sinx · y 1. To antidifferentiate this equation, the first step is to separate the variables. 1/y dy = cos²x · sinx dx 2. Now, taking the integral of the left side is simple. However the right side requires 2. 1/y dy = cos²x · sinx dx the use of substitution. Let u = cos(x) 3. ln|y|+ c = u² du The derivative of cos(x) is sin(x), so Let du = sin(x) dx 4. ln|y|+ c = u³/3 + c 3. Substituting u and du into the equation, we get a simpler expression to 5. ln|y| = (cosx)³/3 + c antidifferentiate. 4. Remember to include +c ! cos³x + c 6. y = e^ 9.Substitute cos(x) back into the equation, 3 and combine the c values onto one side. cos³x c 7. y = e · 10.Isolate y by making the entire left side e^ 3 an exponent of e. cos³x 11.Multiplying two powers of the same y=C e^ 3 base means the exponents are added, so working that fact backwards, you can remove e^c, resulting in a constant C.
22. Slide 22: cos³x 1. To find the value of C, 3 1. v(t) = C e^ simply plug in the cos³(0) coordinates given in the 3 (2)e ⅓ = C e^ question (0, 2e⅓). 2e ⅓ = Ce⅓ 2. This works out to C = 2 C=2 2 . 3 cos³x . 3 v(t) = 2e^ 3. Now just plug in the value for C back into the equation. The equation describingcos³x the imp’s velocity is v(t) = 2e^ 3
23. Slide 23: Reflection Well, it’s done. I must admit, I really procrastinated on this… most of it was done in the past 3 days. That leads me to the first thing I’ve learned. Select deadlines wisely when given the opportunity.. And stop procrastinating.  now, on with the show. The concepts I selected were based on the types of problems we would come across near the ends of units. I was trying to use word problems as they tend to cover several different aspects of a unit and generally also span over other units in some way. I was also looking for things that I needed to brush up on, as the exam is very VERY close now. I think my project covers a fairly good span of the course. It’s pretty easy to do so in this course, as each chapter really builds on the concepts of earlier chapters. I don’t think that’s the case as much in 30S and 40S. As for learning from this assignment, I wouldn’t say it really increased my knowledge a lot, but I think I picked up some details I hadn’t the first time around. I also think that it was a good way to solidify my understanding of specific topics and study for the exam. I’m a bit worried that I’ve accidentally made some glaring errors, and if that’s the case then I guess it wasn’t very valuable. However that would be entirely my fault for not leaving enough time to get feedback from someone.