Questions and Answers #2 Negative Numbers & Coordinate Plane

Questions and Answers Sheet 2 Negative Numbers & Coordinate Plane Question #1 1 Consider the function below. 𝐶(𝑥) = 𝑥 5 (𝑥 + 6) (a) Find the interval of increase. (Enter your answer using interval notation.) Find the interval of decrease. (Enter your answer using interval notation.) b) Find the local minimum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Find the local maximum value(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) (c) Find the inflection points. (x, y)=(?) (smaller x-value) (x, y)=(?) (large x-value) Find the intervals where the graph is concave upward. (Enter your answer using interval notation.) Find the interval where the graph is concave downward. (Enter your answer using interval notation.) Answer: Step 1 Consider the function: 1 𝑓(𝑥) = 𝑥 5 (𝑥 + 6) 1 4 1 a) (1) 𝑓 ′ (𝑥) = 𝜋 −5 (𝑥 + 6) + 𝑥 5 5 1 𝑥+6 5 = 4 +𝑥 5 5𝑥 1 4 = = = + 𝑥+6+5𝑥 5 5 4 5𝑥 5 𝑥+6+5𝑥 4 5𝑥 5 6(𝑥+1) 4 5𝑥 5 For the critical points: f'(x)=0 6(𝑥 + 1) ⇒ =0 4 5𝑥 5 ⇒ 𝑥 + 1 = 0 and 𝑥 ≠ 0 ⇒ 𝑥 = −1 and 𝑥 ≠ 0 The critical points are x=-1 and x=0 Thus, the interval (−∞, −1), (−1,0), (0, ∞) In the interval (−∞, −1): Take x=-2 𝑓 ′ (𝑥) = 6(−2+1) 4 = −0.68 < 0 5(−2)5 ⇒ 𝑓(𝑥) is decreasing in the interval (−∞, −1) In the interval (-1, 0) : Take x=-0.5 𝑓 ′ (𝑥) = 6(−0.5+1) 4 = 1.04 > 0 5(−0.5)5 ⇒ 𝑓(𝑥) is increasing in the interval (-1,0) In the interval (0, ∞): Take x=1 𝑓 ′ (𝑥) = 6(1+1) 4 = 2.4 > 0 5(1)5 ⇒ 𝑓(𝑥) is increasing in the interval (0, ∞) Step 2 b) 𝑓 ′′ (𝑥) = (1) 𝑓 =− ′′ (𝑥) 24 25 6(𝑥+1) 4 5𝑥 5 6 4 = 𝑥 −5 (𝑥 + 1) 5 6 4 5 5 4 6 4 = ⋅ (− ) 𝑥 −5−1 (𝑥 + 1) + 𝑥 −5 9 6 4 5 𝑥 −3 (𝑥 + 1) + 𝑥 −5 5 6 4 4 5 6 4 5 4 = 𝑥 −5 (− 𝑥 −1 (𝑥 + 1) + 1) = 𝑥 −5 (− 5 6 4 − 5 (𝑥 + 1) + 1) 5𝑥 −4𝑥−4+5𝑥 = 𝑥 ( ) 5 5𝑥 6 4 𝑥−4 = 𝑥 −5 ( ) 5 5𝑥 9 6 = 𝑥 −5 (𝑥 − 4) 25 For the infection points: f''(x)=0 6 −9 ⇒ 𝑥 5 (𝑥 − 4) = 0 25 9 ⇒ 𝑥 −5 =≠ 0 and x-4=0 ⇒ 𝑥 ≠ 0 and x=4 Thus, the infection points x=0 and x=4 Local maximum and minimum: 6 4 𝑓 ′′ (0) = (0)−5 (0 + 1) = ∞ 5 ⇒ At x=0 , there is a saddle point. 4 6 𝑓 ′′ (4) = (4)−5 (4 + 1) = 1.979 > 0 5 ⇒ at x=4 , there is a local minimum. 1 𝑓(4) = 45 (4 + 6) = 13.195 Local minimum=13.195 Local maximum=None The infection points are x=0 and x=4 The interval are (−∞, 0), (0,4) and (4, ∞) In the interval (−∞, 0): 𝑓 ′′ (𝑥) = 6 25 9 (−1)−5 (−1 − 4) = 1.2 > 0 f is concave upward In the interval (0, 4) Take x=1 𝑓 ′′ (𝑥) = 6 25 9 (1)−5 (1 − 4) = −0.72 < 0 f is concave downward In the interval (4, ∞): Take x=5 : 𝑓 ′′ (𝑥) = 6 25 9 (5)−5 (5 − 4) = 0.013 > 0 f is concave upward Infection on points: (𝑥, 𝑦) = (0,0) (smaller x value) (𝑥, 𝑦) = (4, 13.195) (larger x value) Concave upward: (−∞, 0) ∪ (4, ∞) Concave downward (0, 4) Question #2 Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of the upper reservoir is 45 m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03𝑚3 /𝑠 , determine mechanical power that is converted to thermal energy during this process due to frictional effects. Answer: First we need to calculate the rate of the mechanical energy increase of the water: △ 𝐸𝑚𝑒𝑐ℎ , in = 𝑚𝑔ℎ = 𝜌𝑉𝑔ℎ 𝑘𝑔 𝑚3 𝑚 = 1000 3 ⋅ 0.03 ⋅ 9.81 2 ⋅ 45𝑚 𝑚 𝑠 𝑠 = 13.2𝑘𝑊 The difference of the invested power and the rate of the energy increase then gives us the power dissipated: △ 𝐸𝑚𝑒𝑐ℎ , lost = 𝑊𝑚𝑒𝑐ℎ ,in −△ 𝐸𝑚𝑒𝑐ℎ , in = (20 − 13.2)𝑘𝑊 = 6.8𝑘𝑊 Result: △ 𝐸𝑚𝑒𝑐ℎ , lost = 6.8𝑘𝑊 Question #3 Find an equation of the tangent plane to the given surface at the specified point. 𝑧 = 2(𝑥 − 1)2 + 6(𝑦 + 3)2 + 4, (3, −2,18) Answer: Find an equation of the tangent plane to the given surface at the specified point. 𝑧 = 2(𝑥 − 1)2 + 6(𝑦 + 3)2 + 4, (3, −2,18) ⇒ 𝑧𝑥 = 4(𝑥 − 1) =4x-4 , 𝑧𝑦 = 12(𝑦 + 3) = 12𝑦 + 36 At the point (3,-2,18): 𝑧𝑥 = 4𝑥 − 4 =4(3)-4 =12-4 =8 𝑧𝑦 = 12𝑦 + 36 =-12(-2)+36 =-24+36 =12 Equation of the tangent plane is, 𝑧 − 𝑧1 = 𝑧𝑥 (𝑥 − 𝑥1 ) + 𝑧𝑦 (𝑦 − 𝑦1 ) z-18=8(x-3)+12(y-(-2)) z-18=8x+12y z=8x+12y+18 Question #4 Find an equation of the tangent plane to the given surface at the specified point. 𝑧 = (𝑥 + 2)2 − 2(𝑦 − 1)2 − 5, (2,3,3) Answer: Equation for tangent plane at (𝑥0 , 𝑦0 , 𝑧0 ) is given by 𝑧 − 𝑧0 = 𝑓𝑥 (𝑥0 , 𝑦0 )(𝑥 − 𝑥0 ) + 𝑓𝑦 (𝑥0 , 𝑦0 )(𝑦 − 𝑦0 ) Given that 𝑓(𝑥, 𝑦) = (𝑥 + 2)2 − 2(𝑦 − 1)2 − 5 Therefore 𝑓𝑥 = 2(𝑥 + 2) 𝑓𝑥 (2,3) = 2(2 + 2) = 8 𝑓𝑦 = −4(𝑦 − 1) 𝑓𝑦 (2,3) = −4(3 − 1) = −8 Equation of the plane at (2,3,3) is 𝑧 − 3 = 𝑓𝑥 (2,3)(𝑥 − 2) + 𝑓𝑦 (2,3)(𝑦 − 3) Substitute 𝑓𝑥 (2,3) = 8, 𝑓𝑦 (2,3) = −8 𝑧 − 3 = 8 ⋅ (𝑥 − 2) − 8 ⋅ (𝑦 − 3) 𝑧 − 3 = 8𝑥 − 16 − 8𝑦 + 24 𝑧 = 8𝑥 − 8𝑦 + 11 Result z=8x-8y+11 Question #5 Find the points on the cone 𝑧 2 = 𝑥 2 + 𝑦 2 that are closest to the point (4,2,0) Answer: Using the distance formula we can say that the distance between a general point (x,y,z) and (4,2,0) is given by 𝐷 = √(𝑥 − 4)2 + (𝑦 − 2)2 + (𝑧 − 0)2 Since the points are on the cone we can replace 𝑧 2 with 𝑥 2 + 𝑦 2 𝐷 = √(𝑥 − 4)2 + (𝑦 − 2)2 + 𝑥 2 + 𝑦 2 Consider the following function 𝑓(𝑥, 𝑦) = (𝑥 − 4)2 + (𝑦 − 2)2 + 𝑥 2 + 𝑦 2 Note that D(x,z) will be minimum when f(x,z) is minimum, so we will work with f(x,z) for conveince To find the minimum value of f(x,y), we will us the second derivative test. Second derivatives test: If the second partial derivatives of f are continuous on a disk containing the critical point (a,b) , and 2 𝐷(𝑎, 𝑏) = 𝑓𝑥𝑥 (𝑎, 𝑏)𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)] Then we can conclude that: 1) D>0 and 𝑓𝑥𝑥 (𝑎, 𝑏) < 0 ⇒ a local minimum 2) D>0 and 𝑓𝑥𝑥 (𝑎, 𝑏) < 0 ⇒ a local maximum 3) 𝐷 < 0 ⇒ no local maximum/minimum We can find the local extrema using the Second Derivatives Tes from the section. Note that 𝑓(𝑥, 𝑦) = (𝑥 − 4)2 + (𝑦 − 2)2 + 𝑥 2 + 𝑦 2 is a polynomial function and hence is continuous everywhere in its domain. Thus the theorem applies. Firstly, we will need the first and second partial derivatives of f: 𝑓𝑥 (𝑥, 𝑦) = 2(𝑥 − 4) + 2𝑥, 𝑓𝑦 (𝑥, 𝑦) = 2(𝑦 − 2) + 2𝑦 ∴ 𝑓𝑥𝑥 (𝑥, 𝑦) = 2(1) + 2(1) = 4𝑓𝑦𝑥 (𝑥, 𝑦) = 0 𝑓𝑥𝑦 (𝑥, 𝑦) = 0𝑓𝑦𝑦 (𝑥, 𝑦) = 2(1) + 2(1) = 4 Now we find the critical points of f. These are the points where both first partial derivatives are 0 at the same time. So let's solve the system of equations 2( x − 4) + 2 x = 0  2( y − 2) + 2 y = 0 Solve the first equation (x-4)+x=0 2x=4 Divide both sides by 2 x=2 Solve the first equation 2(y-2)+2y=0 y-2+y=0 2y=2 Divide both sides by 2 y=1 Therefore, the critical point is (2,1) Now we just need the value 𝐷(𝑎, 𝑏) = 𝑓𝑥𝑥 (𝑎, 𝑏)𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)] 2 and 𝑓𝑥𝑥 at the critical point: 2 𝐷(2,1) = 𝑓𝑥𝑥 (2,1) − [𝑓𝑥𝑦 (2,1)] Since D=16>0 and 𝑓𝑥𝑥 = 4 > 0 , this is a local minimum The z-coordinate of the local minimum is given by 𝑧 2 = 12 + 22 = 5 ⇒ 𝑧 = ±√5 The local minimum is also absolute minimum because the domain is all real numbers and the limits at infinities are positive infinity. Result: (2,1, ±√5) are the closest to the given point Question #6 Find the point(s) on the surface at which the tangent plane is horizontal. 1 1 𝑧 = 𝑥𝑦 + + 𝑥 𝑦 Answer: Find the partial derivatives with respect to x and y and set equal to zero. Solve for x in the partial with respect to y and plug the result back into the partial with respect to y and plug the result back into the partial with respect to x to solve for y, Y cannot equal zero since we cannot have a zero in the denominator so y must equal 1. Plug 1 into the equation for x to find x. 1 1 𝑧 = 𝑥𝑦 + + 𝑥 𝑦 1 =0 𝑥2 1 𝑓𝑦 (𝑥, 𝑦) = 𝑥 − 2 = 0 𝑓𝑥 (𝑥, 𝑦) = 𝑦 − 𝑦 𝑥= 𝑦− 1 𝑦2 1 1 𝑦2 = −𝑦 2 + 𝑦 = 0 y(-y+1)=0 y=1 1 𝑥= 2=1 1 Plug the point (1,1) into z to find the 3rd coordinate. 1 1 𝑧(1,1) = 1 ⋅ 1 + + = 3 1 1 Result: (𝑥, 𝑦, 𝑧) = (1,1,3) Question #7 Find the volume of the solid enclosed by the paraboloid 𝑧 = 2 + 𝑥 2 + (𝑦 − 2)2 and the planes z=1,x=-1y=0, and y=4 Answer: The main problem here is to determine what volume we are calculating. Notice that the solid lies between two surface, one given by the equation 2 + 𝑥 2 + (𝑦 − 2)2 = 𝑧 and the other given by the equation z=1. Hence the idea here is to imagine that the volume of the surface given by the equation 2 + 𝑥 2 + (𝑦 − 2)2 = 𝑧 that is above the xy plane and the given rectangle is being cut off by the plane z=1. By this analogy, we determine that the volume we are seeking is the difference of the two volumes given by the following integrals: 1 4 1 4 ∫−1 ∫0 2 + 𝑥 2 + (𝑦 − 2)2 𝑑𝑦𝑑𝑥 − ∫−1 ∫0 𝑑𝑥𝑑𝑦 We have: 1 4 ∫−1 ∫0 2 + 𝑥 2 + (𝑦 − 2)2 𝑑𝑥𝑑𝑦 − 8 = Result: 88 3 −8= 64 3 64 3 Question #8 Let P(x, y) be a point on the graph of 𝑦 = √𝑥. Express the distance, d, from P to (2, 0) as a function of the point's x-coordinate. Answer: 2 𝑑 = √(𝑥– 2)2 + 𝑦 2 = √(𝑥 − 2)2 + (√𝑥) = √𝑥 2 – 4𝑥 + 4 + 𝑥 = √𝑥 2 – 3𝑥 + 4 Question #9 a) Determine the quadrant in which the angle lies, and (b) list one positive and one negative coterminal angle −5𝜋/6 . Answer: a) Quadrant 3 b) We find coterminals of θ by adding or subtracting integer multiples of 2𝜋: 𝜃 = − 5𝜋/6 + 2𝜋 = 7𝜋/6 - positive 𝜃 = − 5𝜋/6 – 2𝜋 = − 17𝜋/6 - negative Question #10 Find the area of a sector of a circle having radius r and central angle 𝜃. Express answers to the nearest tenth. 𝑟 = 29.2𝑚, 𝜃 = 5𝜋 6 radians Answer: The area A of a sector of a circle of radius r formed by a central angle with radian measure 𝜃 is 1 𝐴 = 𝑟2 𝜃 2 Substitute 𝑟 = 29.2𝑚, 𝜃 = 1 5𝜋 2 6 𝐴 = (29.2𝑚)2 ⋅ Use a calculator 𝐴 ≈ 1116.1𝑚2 5𝜋 6 and solve for A