Q&A #3: Parametric Equations & Polar Coordinates

Questions and Answers Sheet 3 Parametric Equations, Polar Coordinates, and Vector-V Question #1 Express as a trigonometric function of one angle. a) cos 2 sin(−9) − cos 9 sin 2 Find the exact value of the expression. b) sin (arcsin √3 2 + arccos 0) Answer: Step 1 cos 2 sin(−9) − cos 9 sin 2 = cos 2 − sin 𝑞 − cos 9 sin 2 sin(−𝜃) = − sin 𝜃 = − cos 2 sin 9 − cos 9 sin 2 = −(sin 9 cos 2 + cos 9 sin 2) = − sin(9 + 2) = − sin 1 1 ∵ sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵 = sin(−11) ∵ sin(−𝜃) = − sin 𝜃. Step 2 sin [𝑎𝑟𝑐 sin √3 2 √3 + 𝑎𝑟𝑐 cos 0] = sin [sin−1 ( ) + cos −1 (0)] 2 = sin[60∘ + 90∘ ] = sin(150∘ ) = sin(90∘ + 60∘ ) ∵ sin 6 0∘ = = cos 6 0∘ cos 9 0∘ = 0 1 = 2 cos 6 0∘ = √3 2 1 2 Question #2 Use the table of values of f(x, y) to estimate the values of fx(3, 2), fx(3, 2.2), and fxy(3, 2) . y 1.8 2.0 2.2 x 2.5 12.5 10.2 9.3 3.0 18.1 17.5 15.9 3.5 20.0 22.4 26.1 Answer: 𝑓𝑥(3.2) = = (𝑓(3.5, 2)−𝑓(2.5, 5)) (22.4−10.2) 1.0 (3.5−2.5) = 12.2 1.0 =12.2 𝑓𝑥(3,2.2) = = 26.1−9.3 1 (𝑓(3.5,2.2)−𝑓(2.5,2.2)) (3.5−2.5) = 16.8 For fxy(3, 2), we have choices. If we don't care about centering the y, we can use the results already gathered here. Then, 𝑓𝑥𝑦(3.2) = = (16.8−12.2) 2 (𝑓𝑥(3, 2.2)−𝑓𝑥(3.2)) (2.2−2) = 23 If we had wanted to center it, we could have used (𝑓(3.5,2.2) − 𝑓(3.5,1.8) − 𝑓(2.5,2.2) + 𝑓(2.5,1.8)) ((2.2 − 1.8) ⋅ (3.5 − 2.5)) = (26.1−20−9.3+12.5) (4⋅1) = 23.25 Question #3 Find the volume of the solid generated by revolving the standed region about the x-axis. Te volume of the solid is ? cubic units. The equation: 4𝑥 + 3𝑦 = 24 Answer: 4 ⇒ 𝑦 = 8 −( )𝑥 3 1 ⇒ 𝑦 = 4 × (2 − ( ) 𝑥) 3 for x intercept, y=0 1 0 = 4 × (2 − ( ) 𝑥) 3 1 ( )𝑥 = 2 3 x=6 6 Volume generated by rotating the given region about x axis by washer method 𝑣 = ∫0 𝜋 [4 × (2 − 2 1 ( ) 𝑥)] 𝑑𝑥 3 6 1 1 3 3 2 𝑣 = ∫0 16𝜋 [22 − 2 × 2 ( ) 𝑥 + (( ) 𝑥) ] 𝑑𝑥 6 4 1 𝑣 = ∫ 16𝜋 [4 − ( ) 𝑥 + ( ) 𝑥 2 ] 𝑑𝑥 3 9 0 6 4 1 1 3 2 2 2 9 1 𝑣 = ∫0 16𝜋 [4𝑥 − ( ) ( ) 𝑥 2 + ( ) ( ) 𝑥 3 ] 𝑣= 6 ∫0 16𝜋 [4𝑥 1 3 3 − ( )𝑥 + ( )𝑥 ] 3 27 2 2 1 2 1 𝑣 = 16𝜋 [4 × 6 − ( ) 6 + ( ) 63 ] − 16𝜋 [4 × 0 − ( ) 02 + ( ) 03 ] 3 27 3 27 36 216 𝑣 = 16𝜋 [4 × 6 − (2 × ) + ( )] 3 27 𝑣 = 16𝜋[24 − 24 + 8] 𝑣 = 128𝜋 Volume = 128𝜋 = 402 Question #4 Evaluate the double integral 𝑦 2 𝑑𝐴, D is the triangular region with vertices (0, 1), (1,2), (4,1) Answer: Step 1 Let's first define the lines that bound our region: y=1 For (0, 1) and (1, 2): 2−1 (𝑥 − 0) ⇒ 𝑦 = 𝑥 + 1 ⇒ 𝑥 = 𝑦 − 1 𝑦−1 = 1−0 For (1, 2) and (4, 1): 1−2 −1 7 (𝑥 − 1) ⇒ 𝑦 = 𝑥 + ⇒ 𝑥 = 7 − 3𝑦 𝑦−2 = 4−1 3 3 Step 2 Now we are ready to evaluate the double integral: 2 7−3𝑦 ∫ ∫𝐷 𝑦 2 𝑑𝑥𝑑𝑦 = ∫1 ∫𝑦−1 𝑦 2 𝑑𝑥𝑑𝑦 2 7−3𝑦 2 = ∫1 [𝑥𝑦 2 ]𝑦−1 𝑑𝑦 = ∫1 [7𝑦 2 − 3𝑦 3 − 𝑦 3 + 𝑦 2 ]𝑑𝑦 8 2 3 11 1 8 3 3 2 = ∫1 8𝑦 2 − 4𝑦 3 𝑑𝑦 = [ 𝑦 3 − 𝑦 4 ] =[ 64 3 − 16 − + 1] = Thus we conclude that our result is 11 3 Question #5 Find an equation of the plane tangent to the following surface at the given point. 7𝑥𝑦 + 𝑦𝑧 + 4𝑥𝑧 − 48 = 0; (2,2,2) Answer: 𝑓(𝑥, 𝑦, 𝑧) = 7𝑥𝑦 + 𝑦𝑧 + 4𝑥𝑧 − 48 = 0(2,2,2) Tangent direction is (𝑃𝑥 , 𝑃𝑦 , 𝑃𝑧 ) 𝑃𝑥 = 7𝑦 + 𝑦𝑧, 𝑃𝑦 = 7𝑥 + 𝑧, 𝑃𝑧 = 𝑦 + 4𝑥 at (2,2,2) 𝑃𝑥 = 22, 𝑃𝑦 = 16, 𝑃𝑧 = 10 (22,16,10) ⇒ 𝑥(𝑡) = (2,2,2) + 𝑡(22,16,10) ⇒ 22(𝑥 − 𝑧) + 16(𝑦 − 𝑧) + 10(𝑧 − 𝑧) = 0 22𝑥 + 16𝑦 + 10𝑧 − 96 = 0 Question #6 Find the volume of the solid that is enclosed by the cone 𝑧 = (𝑥 2 + 𝑦 2 )1/2 and the sphere 𝑥 2 + 𝑦 2 + 𝑧 2 = 2 Answer: Convert the equation od the cone to cylindrical coordinates 𝑧 = √𝑥 2 + 𝑦 2 = √𝑟 2 = 𝑟 Do the same for the equation of the sphere. 𝑥 2 + 𝑦 2 + 𝑧 2 = 2 → 𝑧 2 = 2 − 𝑟 2 , 𝑧 = √2 − 𝑟 2 Find the intersection of the sphere the cone by setting the equations equal to each other and solving for r. This value is the maximum value for r in our integral. 𝑟 = √2 − 𝑟 2 → 2𝑟 2 = 2, 𝑟 = 1 Set up the integral. Since the region is spherical of the x/y axis, we integrate from 0 to 2 pi. The equation of the sphere defines the top of the region, and the equation of be cone defines the bottom Integrate with respect to z 2𝜋 ∫ 1 √ 2 𝑧= 2−𝑟 ∫ 𝑟 ⋅ 𝑧|𝑧=𝑟 𝑑𝑟𝑑𝜃 0 0 2𝜋 1 ∫0 ∫0 𝑟√2 − 𝑟 2 − 𝑟 2 𝑑𝑟𝑑 𝜃 Break the inner integral into separate pieces so that we can perform a u-substitution on the left portion. 2𝜋 ∫ 0 1 1 [∫ 𝑟√2 − 𝑟 2 𝑑𝑟 − ∫ 𝑟 2 𝑑𝑟] 𝑑𝜃. 0 0 𝑢 = 2 − 𝑟 2 , 𝑑𝑢 = −2𝑟𝑑𝑟 Perform the substitution and simplify. 2𝜋 1 1 −1 ∫ [∫ √𝑢 𝑑𝑢 − ∫ 𝑟 2 𝑑𝑟] 𝑑𝜃 0 2 2 0 2𝜋 21 1 ∫0 [∫1 2 √𝑢 𝑑𝑢 − ∫0 𝑟 2 𝑑𝑟] 𝑑𝜃 . Integrate with respect to u and r, and simplify 2𝜋 1 1 −1 ∫ [∫ √𝑢 𝑑𝑢 − ∫ 𝑟 2 𝑑𝑟] 𝑑𝜃 0 2 2 0 2𝜋 2 1 1 ∫ [∫ √𝑢 𝑑𝑢 − ∫ 𝑟 2 𝑑𝑟] 𝑑𝜃 0 1 2 0 2𝜋 2 ∫ (√2 − 1)𝑑𝜃. 0 3 Integrate with respect to theta 2 4𝜋 (√2 − 1) ⋅ 𝜃|𝜃=2𝜋 (√2 − 1) 𝜃=0 = 3 3 Result: 4𝜋 (√2 − 1) 3 Question #7 Use set theoretic or vector notation or both to describe the points that lie in the given configurations. The line passing through (0,2,1) in the direction of 2i-k Answer: It is given that point 𝛼 = (0,2,1) and 𝑣1 = (2,0, −1) The equation of the line l through the tip of 𝛼 and pointing in the direction of the vector v is 𝑙(𝑡) = 𝑎 + 𝑡𝑣, where the parameter t takes on all real values. Thus we can write l=(0,2,1)+(2,0,-1)t l=(2t,2,1-t) Result: l=(2t,2,1-t) where 𝑡 ∈ 𝑅 Question #8 Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. 𝑥 = 1 + 2√𝑡, 𝑦 = 𝑡 3 − 𝑡, 𝑧 = 𝑡 3 + 𝑡; (3,0,2) Answer: We're given parametric equation 𝑥 = 1 + 2√𝑡, 𝑦 = 𝑡 3 − 𝑡 and 𝑧 = 𝑡 3 + 𝑡 and we're asked to solve for the parametric equations of the tangent line to the curve at the point (3,0,2) Given parametric equations, we know that the vector equation tr(t) is equal to 𝑟(𝑡) =< 1 + 2√𝑡, 𝑡 3 − 𝑡, 𝑡 3 + 𝑡 > Solve for r'(t) by differentiating each of the components of r(t) with respect to t 1 𝑟 ′ (𝑡) =< , 3𝑡 2 − 1,3𝑡 2 + 1 > √𝑡 The parameter value corresponding to (3,0,2) is t=1. Plug in t=1 into r'(t) to solve for r'(1) 1 𝑟 ′ (1) =< , 3(1)2 − 1,3(1)2 + 1 > √1 =<1,2,4> Recall from the textbook that parametric equations for a line through the point (𝑥0 , 𝑦0 , 𝑧0 ) and parallel to the direction vector are 𝑥 = 𝑥0 + 𝑎𝑡 𝑦 = 𝑦0 + 𝑏𝑡𝑧 = 𝑧0 + 𝑐𝑡 Substitute (𝑥0 , 𝑦0 , 𝑧0 ) = (3,0,2) and =<1,2,4> into x, y and z, respectively to solve for the parametric equations of the tangent line to the curve 𝑥 = 3 + (1)𝑡 = 3 + 𝑡, 𝑦 = (0) + (2)𝑡 = 2𝑡, 𝑧 = (2) + (4)𝑡 = 2 + 4𝑡 Question #9 Find parametric equations and symmetric equations for the line. 1 1 The line through (-6, 2, 3) and parallel to the line = =𝑧+1 2𝑥 3𝑦 Answer: Rationalization. Given a dispacement vector v= which is the difference of the initial position vector 𝑟0 and final position vector r as well as the value of an initial position vector 𝑟0 , we can acquire the parametric equations of a line:  x = x0 + at   y = y0 + bt  z = z + ct 0  From this parametric equation, we can acquire the parametric equations of L as follows: 𝑥−𝑥0 𝑦−𝑦0 𝑧−𝑧0 = = 𝑎 𝑏 𝑐 Deriving the parametric equations: finding displacement vector. 1 1 𝑥+ 𝑦 =𝑧+1 2 3 This is already a symmetric equation of a line parallel to the given line that passes through (-6,2,3) with position vector 𝑟0 =< −6,2,3 > . We can rewrite this as a parametric equation by equating to t:  1 t = 2 x  x = 2t  1 1  1  t = x = u = z + 1  t = y   y = 3t 2 3  3 z = t −1 t = z + 1    Deriving the symmetric equations. The symmetric equations are the three-fold equaliy we acquire between x, y and z by eliminating t. It was already mentioned in the chapter that a shorthand formula for this from the parametric equations is: 𝑥−𝑥0 𝑦−𝑦0 𝑧−𝑧0 = = 𝑎 𝑏 𝑐 Substituting calues, we get: 𝑥+6 𝑦−2 𝑧−3 𝑥+6 𝑦−2 = = = = =𝑧−3 2 3 1 2 3 Question #10 If 𝑥𝑦 + 8𝑒 𝑦 = 8𝑒, find the value of y'' at the point where x = 0. Answer: Given, 𝑥𝑦 + 8𝑒 𝑦 = 8𝑒 To find the value of y'' at the point x=0. We have, 𝑥𝑦 + 8𝑒 𝑦 = 8𝑒(1) Differentiating with respect to x, ⇒ 𝑥𝑦 ′ + 𝑦 + 8𝑒 𝑦 𝑦 ′ = 0(2) Differentiating again with respect to x, ′ ′ ⇒ 𝑥𝑦 ′′ + 𝑦 ′ + 𝑦 ′ + 8𝑒 𝑦 𝑦 ′′ + 8𝑒 𝑦 𝑦 = 0 𝑦 ′′ ′ 𝑦 ′ ⇒ (𝑥 + 88𝑒 )𝑦 + 2𝑦 + 8𝑒 (𝑦 )2 = 0(3) Put x=0 in (1),(2) and (3), 0 + 8𝑒 𝑦 = 8𝑒 ⇒ 𝑒𝑦 = 𝑒 ∴𝑦=1 By (2), ⇒ 0 + 1 + 8𝑒𝑦 ′ = 0 ⇒ 8𝑒𝑦 ′ = −1 ∴ 𝑦′ = −1 8𝑒 By (3), 1 1 8𝑒 8𝑒 2 ⇒ (0 + 8𝑒)𝑦 ′′ + 2 (− ) + 8𝑒 (− ) = 0 ⇒ 8𝑒𝑦 ′′ − ⇒ 8𝑒𝑦 ′′ − ∴ 𝑦 ′′ = 2 8𝑒 1 8𝑒 1 8𝑒(8𝑒) + 1 8𝑒 =0 =0 = 1 64𝑒 2