4. 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
5. 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.
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. 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
9. 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
10. 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
11. 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.
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. 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
17. 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
18. 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
22. 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
