Questions and Answers #1 Negative Numbers & Coordinate Plane

Questions and Answers Sheet 1 Negative Numbers & Coordinate Plane Question #1 Find an equation of the tangent line to the curve at the given point. 𝑦 = √𝑥, (81,9) Answer: 𝑦 = √𝑥, (81,9) 𝑦 = √𝑥 Differentiating with respect x, we get 𝑑𝑦 1 = 𝑑𝑥 2√𝑥 𝑑𝑦 𝑑𝑥 = 1 2√81 = 1 2(9) = 1 18 The equation of tangent with slope 1 (𝑥 − 81) 𝑦−9 = 18 1 81 𝑦−9 = 𝑥− 18 18 1 9 𝑦−9 = 𝑥− 18 2 1 9 𝑦= 𝑥− +9 18 2 1 −9 + 18 𝑦= 𝑥+ 18 2 1 9 𝑦= 𝑥+ 18 2 1 18 and the point (81,9) is Question #2 Find the area of the region enclosed by the inner loop of the curve. 𝑟 = 2 + 4 sin θ Answer: Find the interval in which the inner loop exists. For this we take 𝑟 = 0 ⇒ 2 + 4 sin θ = 0 1 ⇒ sin θ = − 2 7π 11π ⇒θ= , 6 6 7π 11π So inner loop lies between the lines θ = and θ = 6 6 Since curve is symmetric about the vertical polar axis. So we can find the area of the loop by making 7π 3π double the area between θ = and θ = 6 2 3π/2 1 Then Area of the loop is 𝐴 = 2 ∫7π/6 𝑟 2 𝑑θ 2 Therefore 3π/2 (2 + 4 sin θ)2 𝑑θ 𝐴=∫ 7π/6 3π/2 𝐴 = 4∫ 7π/6 (1 + 2 sin θ)2 𝑑θ 3π/2 (1 + 2(1 − cos 2 θ) + 4 sin θ)𝑑θ = 4∫ 7π/6 3π/2 (3 − 2 cos 2 θ + 4 sin θ)𝑑θ = 4∫ 7π/6 3𝜋/2 = 4[3𝜃 − 𝑠𝑖 𝑛 2 𝜃 + 4(−𝑐𝑜 𝑠 𝜃)]7𝜋/6 9π = 4[ 2 − 7π 2 3 − √3] 2 𝐴 = 4π − 6√3 Question #3 Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. 𝑦 = 𝑥 2 , 𝑥 = 𝑦 2 ; about 𝑦 = 1 Answer: Question #4 Find the point at which the line intersects the given plane. 𝑥 = 2 − 2𝑡 , 𝑦 = 3𝑡, 𝑧 = 1 + 𝑡; 𝑥 + 2𝑦 − 𝑧 = 7 Answer: Substitute 𝑥 = 2 − 2𝑡 , 𝑦 = 3𝑡 , 𝑧 = 1 + 𝑡 𝑖𝑛 𝑥 + 2𝑦 − 𝑧 = 7 (2 − 2𝑡) + 2(3𝑡) − (1 + 𝑡) = 7 2 − 2𝑡 + 6𝑡 − 1 − 𝑡 = 7 3𝑡 + 1 = 7 3𝑡 = 6 𝑡=2 Substitute 𝑡 = 2 in equation of the line to get the point 𝑥 = 2− 2 ⋅ 2 ,𝑦 = 3 ⋅ 2 ,𝑧 = 1 + 2 𝑥 = −2, 𝑦 = 6 , 𝑧 = 3 (−2,6,3) Question #5 Write the equations in cylindrical coordinates. 𝑥 2 − 𝑥 + 𝑦 2 + 𝑧 2 = 1 Answer: 1) Let's use the following formulas to find the cylindrical coordinates: 𝑟2 = 𝑥 2 + 𝑦2 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 sin 𝜃 𝑧=𝑧 2) Now let's rearrange the given expression: 𝑥 2 + 𝑦2 − 𝑥 + 𝑧2 = 1 3)We can now substitute the formulas from 1) and get our solution: 𝑟 2 − (𝑟 cos 𝜃) + 𝑧 2 = 1 Result 𝑟 2 − (𝑟 cos 𝜃) + 𝑧 2 = 1 Question #6 Find the minimum and maximum values of 𝑦 = √6𝜃 − √3 sec 𝜃 on the interval pi [0, ] 3 Answer: [𝑦 = √6𝜃 − √6 sec 𝜃 on the interval [ 0, 𝜋/3]) [𝑓(0) = −√(6) [𝑓(𝑝𝑖/3) = √(6)(𝜋/3) − √(3) sec(𝜋/3) = √(6)(𝜋/3) − √(3)(2) = −0.899 [𝑓 ′ (𝑥) = √(6) − √(3) sec(𝜃) tan(𝜃) = 0 [√(6) = √(3) sec(𝜃) tan(𝜃) [sec(𝜃) tan(𝜃) = √(6)/√(3) [ 𝜃 = 𝜋/4 [𝑓(𝜋/4) = √(6)(𝜋/4) − √(3) sec(𝜋/4) = −0.52566 we have: [𝑓(0) = −√(6) = −2.4495 [𝑓(𝜋/3) = −0.899 [𝑓(𝜋/4) = −0.52566 minimum = [−√(6) (occurs at x=0) maximum =[ -0.52566(occurs at 𝑥 = 𝜋/4)] Question #7 Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola. (Round your answers to the nearest hundredth.) 𝑦 = 6 − 𝑥2 Answer: Area of the rectangle is A=xy Substitue 𝑦 = 6 − 𝑥2 in the formula of the area. 𝐴 = 𝑥(6 − 𝑥 2 ) = 6𝑥 − 𝑥 3 Diferentiate area with respect to x and equate to 0. 𝑑 𝐴′ = (6𝑥 − 𝑥 3 ) 𝑑𝑥 = 6 − 3𝑥 2 6 − 3𝑥 2 = 0 3𝑥 2 = 6 𝑥 = ±√2 Substitute 𝑥 = ±√2 in 𝑦 = 6 − 𝑥 2 to get value of height, 𝑦 = 6 − 𝑥2 2 = 6 − (√2) =4 Hence. the dimensions of rectangle are, 2√2 × 4. Question #8 Find the area of the surface. The part of the surface 2𝑦 + 4𝑧 − 𝑥 2 = 5 that lies above the triangle with vertices (0, 0), (2, 0), and (2, 4) Answer: Question #9 Find then slope and y-intercept of the line. Graph the line. 5x+y=5 Answer: Let's rewrite it as an equation with a slope. Slope equation type: y = mx + b , where equals the slope and b equals the y-intercept of the line. Subtract 5x from both sides of the equation. y=5-5x Replace 5 and -5x. Let's use the equation for a bunch of straight lines passing through a given point in order to find the slope and the point of intersection with the y-axis. Find the values of m and b using the form y=mx+b m=-5 b=5 The slope of the line is m, and the y-intercept is b. Slope: -5 y-intercept: (0, 5) Question #10 Find an equation of the tangent plane to the given surface at the specified point. 𝑧 = 3𝑦 2 − 2𝑥 2 + 𝑥, (2, −1, −3) Answer: Solve tfor the partial derivative of f with respect to x 𝑓𝑥 (𝑥, 𝑦) = −4𝑥 + 1 𝑓𝑦 (𝑥, 𝑦) = 6𝑦 At the point (2,1) evaluate 𝑓𝑥 = (2, −1) = −7 At the point (2,-1) evaluate 𝑓𝑦 (2, −1) = −6 Frame the equation of the tangent plane 𝑧 + 3 = 𝑓𝑥 (2, −1)[𝑥 − 2] + 𝑓𝑦 (2, −1)[𝑦 + 1} Plug in respective values 𝑧 + 3 = −7𝑥 + 14 − 6𝑦 − 6 Simplify −7𝑥 − 6𝑦 + 5 = 𝑧