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?
1. To solve this problem, you must first find an equation to relate x, y, and z. x ² + y ² = z ² 2. You can then use this to find the distance z. 6 ² + 2 ² = z ² z = 2 √10 km Negative, because it’s going towards the UFO, thus decreasing the distance. y = 2km dy/dt = -144km/h x = 6km dx/dt = ? z = 2 √10 km dz/dt = 2016/ √10 km/h
3. The next step is to get an equation that relates the rates of change of the three distances. You can do this by implicit differentiation, and by using the chain 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 y and z.) Now you have an expression relating all the velocities, and all you have to do is plug in the values given in the problem to solve for the velocity of the UFO (dx/dt). x ² + y ² = z ² 2 x(dx/dt) + 2 y(dy/dt) = 2 z(dz/dt) x(dx/dt) + y(dy/dt) = z(dz/dt) (6)(dx/dt) + (2)(-144) = (2 √10 )(2016/ √10) 6(dx/dt) = 4320 dx/dt = 720 km/h Answer: The UFO is travelling East at a velocity of 720 km/h.
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
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. In this case, the rectangle is symmetrical on either side of the y-axis. If we call the distance from x = 0 to one of the vertices ‘x’, the entire base can be written as ‘2x’. The height ‘y’ of the rectangle is the distance from the top function to the bottom function . Written in terms of x, it is a(x) - b(x) . 3. The area of the rectangle can be written as: A(x) = base * height = 2x[ (-1/2x²+2) - (2x-4) ] =2x(-1/2x²-2x+6) A(x) = -x³-4x²+12x
Solution A We are looking for the maximum area. One way to do this is using the derivative of A(x). A(x) = -x³-4x²+12x A’(x) = -3x²-8x+12 x = -b ± √(b²-4ac) 2a x = -(-8) ± √((-8)²-4(-3)(12)) 2(-3) x = 8 ± √(64+144) -6 x = 8 ± 4 √13 -6 Use the quadratic formula to find the values of x where A’(x)=0. The maximum value A(x) will be found at one of these critical numbers. x = 4 + 2 √13 x = 4 - 2 √13 3 3
Solution A A maximum on the parent function is denoted by a zero at some value ‘x=a’ on the derivative, where when x<a, f’(x) >0, and when x>a, f’(x)<0. x = 4 + 2 √13 x = 4 - 2 √13 3 3 ≈ -3.7370 ≈ 1.0704 Now it is important to remember the context of the problem. We are looking for an area, so the value of x (which describes the length of the sides) can’t be negative. Just to confirm that the remaining zero is a maximum, we do a line test. There is a maximum on the parent function at x = 4 - 2 √13 3
Solution A We are answering the question ‘what is the greatest possible area of the rectangle?’ What we have is the value of x at which this maximum is found. To find the final answer, substitute that x-value back into the equation that describes the area of the rectangle. A(x) = -x³-4x²+12x = -( 4 - 2 √13 ) ³ -4( 4 - 2 √13 ) ²+12 ( 4 - 2 √13 ) 3 3 3 = 7.0354u² x = 4 - 2 √13 3 The maximum possible area of the rectangle is 7.0354u².
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.
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.
f(x)=x²+2 g(x)=x+4 x²+2 = x+4 x²-x-2=0 x = -1 , 2 1. Find out where the two functions intersect, to determine the intervals on which you must integrate. In this case, as you are only looking for the area from [0,2], the intersection at x=2 is the only one that’s necessary. Now you need an equation describing the area of the washers. The area of a circle is π r². The area of the washers is the total area of the circle (with radius equal to the upper function) minus the area of 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]. V₁ = π r² - π r² = π ( x+4 )² - ( x²+2 )² ₂ ³ V ₂ = π ( x²+2 )² - ( x+4 )² ₀ ² ₀ ²
Now solve the integrals: V= π ( x+4 )² - ( x²+2 )² = π -x⁴-3x²+8x+12 = π [-x⁵/5-x³+4x²+12x]² = π (128/5 - 0) =128 π /5u³ ₀ V= π ( x²+2 )² - ( x+4 )² = π x⁴+3x²-8x-12 = π [x⁵/5+x³-4x²-12x]³ = π [18/5 – (-128/5)] =146 π /5u³ ₂ ³ ₂ ³ ₂ The total volume of the solid is the sum of the two volumes. = 128 π /5u³ + 146 π /5u³ = 274 π /5u³ The volume of the solid is 274 π /5u³. ₀ ² ₀ ²
Solution b 1. Your calculator can be used to find the volume of the solid as well. Recall that the volume of the first segment of the solid is expressed by the equation shown. 2. Plug the equation describing the area of the washers into Y1. 3. Press 2 nd 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. 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. Repeat steps 1-4 using the equation for the volume of the green segment. 6. Add the two volumes for your final answer! ( x+4 )² - ( x²+2 )² V= π ( x+4 )² - ( x²+2 )² ₀ ² 1. 2. 3. 4. 25.6 25.6 π u³ 5. 29.2 π u³ 6. 54.8 π u³
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.
dy/dx = cos²x · sinx · y 1/y dy = cos²x · sinx dx 1/y dy = cos²x · sinx dx ln|y|+ c = u ² du ln|y|+ c = u ³/3 + c ln|y| = ( cosx )³/3 + c y = e^ y = e · e^ y = C e^ cos³x + c 3 c cos³x 3 cos³x 3
1. To antidifferentiate this equation, the first step is to separate the variables.
2. Now, taking the integral of the left side is simple. However the right side requires the use of substitution.
Let u = cos(x)
The derivative of cos(x) is sin(x), so
Let du = sin(x) dx
3. Substituting u and du into the equation, we get a simpler expression to antidifferentiate.
4. Remember to include +c !
Substitute cos(x) back into the equation, and combine the c values onto one side.
Isolate y by making the entire left side an exponent of e.
Multiplying two powers of the same base means the exponents are added, so working that fact backwards, you can remove e^c, resulting in a constant C.
1. 2. 3. 4. 5. 6. 7.
v(t) = C e^ (2)e ⅓ = C e^ 2e ⅓ = Ce⅓ C = 2 v(t) = 2e^ cos³x 3 1. To find the value of C, simply plug in the coordinates given in the question (0, 2e⅓). 2. This works out to C = 2 3. Now just plug in the value for C back into the equation. cos³(0) 3 cos³x 3 1. 2. 3. The equation describing the imp’s velocity is v(t) = 2e^ cos³x 3
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.