Questions and Answers #1 Vectors and Spaces

Questions and Answers Sheet 1 Vectors and Spaces Question #1 Differentiate. 𝑓(𝜃) = 𝜃 cos 𝜃 sin 𝜃. Answer: Question #2 Find T (t), N(t), 𝑎 𝑇 , and 𝑎𝑁 at the given time t for the plane curver r(t). 𝑟(𝑡) = 𝑡 2 𝑖 + 2𝑡𝑗, 𝑡 = 1 Answer: Question #3 Black vultures excel at gliding flight; they can move long distances through the air without flapping their wings while undergoing only a modest drop in height. A vulture in a typical glide in still air moves along a path tipped 3.5∘ below the horizontal. If the vulture moves a horizontal distance of 100 m, how much height does it lose? Answer: Question #4 If a vector has direction angles 𝛼 = Answer: 𝜋 4 and 𝛽 = 𝜋 3 , find the third direction angle 𝛾. Question #5 Find the vectors T, N, and B at the given point. 2 16 3 3 𝑟(𝑡) =< 𝑡 2 , 𝑡 3 , 𝑡 > and point < 4, − Answer: , −2 > Question #6 Determine whether the given set S is a subspace of the vector space V. A. 𝑉 = 𝑃5 , and S is the subset of 𝑃5 consisting of those polynomials satisfying 𝑝(1) > 𝑝(0). B. 𝑉 = 𝑅3 , and S is the set of vectors (𝑥1 , 𝑥2 , 𝑥3 ) in V satisfying 𝑥1 − 6𝑥2 + 𝑥3 = 5. C. 𝑉 = 𝑅𝑛 , and S is the set of solutions to the homogeneous linear system Ax=0 where A is a fixed 𝑚 × 𝑛 matrix. D. 𝑉 = 𝐶 2 (𝐼), and S is the subset of V consisting of those functions satisfying the differential equation 𝑦 ″ − 4𝑦 ′ + 3𝑦 = 0. E. V is the vector space of all real-valued functions defined on the interval [a,b], and S is the subset of V consisting of those functions satisfying f(a)=5. F. 𝑉 = 𝑃𝑛 , and S is the subset of 𝑃𝑛 consisting of those polynomials satisfying 𝑝(0) = 0. G. 𝑉 = 𝑀𝑛 (𝑅) , and S is the subset of all symmetric matrices Answer: Question #7 Find the curvature of 𝑟(𝑡) =< 𝑡, 𝑡 2 , 𝑡 3 > at the point (1, 1, 1) Answer: Step 1 We have to find the curvature of the vector function 𝑟(𝑡) =< 𝑡, 𝑡 2 , 𝑡 3 > at point P=<1,1,1> Now, let us find the value for t for which r(t)=P, clearly we get t=1 We know that the formula for curvature of a curve is, 𝑘= |𝑟 ′ (𝑡)×𝑟 ′′ (𝑡)| |𝑟 ′ (𝑡)|3 Now , let us differential the given vector equation component-wise to find the first derivative r'(t) and r''(t) , we get 𝑟 ′ (𝑡) =< 1,2𝑡, 3𝑡 2 > 𝑟 ′′ (𝑡) =< 0,2,6𝑡 > Using the above derivative equations , we could find their cross product , that is 𝑟 ′ (𝑡) × 𝑟 ′′ (𝑡) =< 6𝑡 2 , −6𝑡, 2 > Thus ,we could find the magnitude , that is, |𝑟 ′ (𝑡) × 𝑟 ′′ (𝑡)| = √(6𝑡 2 )2 + (−6𝑡)2 + 22 = √36𝑡 4 + 36𝑡 2 + 4 Step 2 Similarly , let us find the magnitude of r'(t) and cube it, we get 3 |𝑟 ′ (𝑡)| = √1 + 4𝑡 2 + 9𝑡 4 ⇒ |𝑟 ′ (𝑡)|3 = (√1 + 4𝑡 2 + 9𝑡 4 ) Put the components together and find the curvature using formula stated above , that is, 𝑘= 𝑘= |𝑟 ′ (𝑡)×𝑟 ′′ (𝑡)| |𝑟 ′ (𝑡)|3 √36𝑡 4 +36𝑡 2 +4 3 (1+4𝑡 2 +9𝑡 4 )2 Apply t=1 as calculated earlier , we get, 𝑘= 𝑘= 𝑘= 𝑘= 𝑘= √36⋅14 +36⋅12 +4 3 (1+4⋅12 +9⋅14 )2 √36+36+4 3 (1+4+9)2 76 3 (14)2 76 14⋅√14 38 7√14 𝑘 ≈ 1.45 The approximate value of the curvature for given curve is 0.17 Question #8 Find a vector of magnitude 3 in the direction opposite to the direction of 1 1 1 2 2 2 𝑣 = ( )𝑖 − ( )𝑗 − ( )𝑘 Answer: Finding a vector of magnitude 3 in the direction opposite to the vector 1 1 1 2 2 2 𝑣 = ( )𝑖 − ( )𝑗 − ( )𝑘 In order to get the opposite direction of the vector v we need to get the direction of v, then we will reverse the direction by multiplying the direction of v by -1 Hence the direction we are looking for or the unit vector we are looking for is −𝑣 𝑢̂ = |𝑣| We will calculate it as follows −𝑣 𝑢̂ = |𝑣| ⇒ 𝑢̂ = ⇒ 𝑢̂ = 1 1 1 2 2 2 1 1 1 |( 𝑖− 𝑗− 𝑘)| 2 2 2 −1 1 1 ( 𝑖+ 𝑗+ 𝑘) 2 2 2 1 2 −1 2 −1 2 −( 𝑖− 𝑗− 𝑘) √( ) +( ⇒ 𝑢̂ = ⇒ 𝑢̂ = ⇒ 𝑢̂ = ⇒ 𝑢̂ = ) +( 2 2 −1 1 1 ( 𝑖+ 𝑗+ 𝑘) 2 2 2 1 1 1 √ + + 4 4 4 −1 1 1 ( 𝑖+ 𝑗+ 𝑘) 2 2 2 3 √ 4 −1 1 1 ( 𝑖+ 𝑗+ 𝑘) 2 2 2 √3 2 2 ) (−𝑖+𝑗+𝑘) √3 Finally to get the needed vector we will multiply the unit vector we got by 3 (the given magnitude) The needed vector= 3 (−𝑖+𝑗+𝑘) √3 The needed vector= √3(−𝑖 + 𝑗 + 𝑘) Question #9 Find the vector x determined by the given coordinate vector [𝑥]ℬ and the given basis  4 6  8 = {  ,  },[ x] =    5  7   −5 Answer: We calculate that  4  6   2 x = 8   + (−5)   =   5 7   5  Question #10 Determine by inspection whether the vectors are linearly independent. Justify each answer. 5  2 1  −1 1 , 8  , 3 ,  7          Answer: The given vectors are 5  2 1  −1 1 , 8  , 3 ,  7          From Theorem 8, the set of vectors are linearly dependent because here we have four-vector and two entries. Hence, the given set is linearly dependent. Result Linearly dependent. Question #11 Find the vectors T, N , and B at the given point. 𝑟(𝑡) =< cos 𝑡 , sin 𝑡 , ln cos 𝑡 >, (1,0,0) Answer: Step 1 T(t) is the Tangent to the Vector Curve. 𝑟 ′ (𝑡) 𝑇(𝑡) = ′ |𝑟 (𝑡)| 𝑟 ′ (𝑡) =< − sin(𝑡) , cos(𝑡) , − tan(𝑡) > |𝑟 ′ (𝑡)| = √1 + tan2 (𝑡) = √sec 2 (𝑡) = sec(𝑡) The point at which we need to find T,B,N is (1,0,0), Hence t=0 0 1 −0 𝑇(0) =< − , , ≥< 0,1,0 > 1 1 1 Step 2 N is the Normal to the vector curve 𝑇 ′(𝑡) 𝑁 = |𝑇 ′(𝑡)| 𝑇(𝑡) =< − sin(𝑡) cos(𝑡) , cos 2(𝑡) , − sin(𝑡) > 𝑇 ′ (𝑡) =< − cos 2(𝑡) + sin2 (𝑡) >, −2 cos(𝑡) sin(𝑡) , − cos(𝑡) > 𝑇 ′ (𝑡) =< − cos 2 𝑡, − sin 2 𝑡, − cos(𝑡) > |𝑇 ′ (𝑡)| = √1 + cos 2 (𝑡) < −1,0, −1 > 1 1 𝑁(0) = =< − , 0, − > √1 + 1 √2 √2 Step 3 𝐵 = 𝑇 × 𝑁 =< 0,1,0 >×< − 1 √2 , 0, − 1 √2 ≥< − 1 √2 , 0, 1 √2 > Question #12 Find the angle 𝜃 between the vectors (a) in radians and (b) in degrees. 𝑢 = 3𝑖 + 4𝑗 and 𝑣 = −2𝑗 + 3𝑘 Answer: 𝑢 = 3𝑖 + 4𝑗 =< 3,4,0 > 𝑣 = −2𝑗 + 3𝑘 =< 0, −2,3 > 𝑢 ⋅ 𝑣 = ||𝑢|| ⋅ ||𝑣|| ⋅ cos 𝜃 𝑢⋅𝑣 ⇒ cos 𝜃 = ||𝑢||⋅||𝑣|| cos 𝜃 = 3 ⋅ 0 + 4 ⋅ (−2) + 0 ⋅ 3 𝜃 = 116.5∘ Result 116.5∘ 5 ⋅ √13 =− 8 5√13 ≈ −0.4437 Question #13 Find the volume of the solid in the first octant bounded by the cylinder 𝑧 = 16 − 𝑥 2 and the plane y=5 Answer: To solve this, it is inportant to set up the boundaries of double integral correctly. First, notice that the cylinder we are given intersects the zy plane at the line given by the equation x = 4. Hence our boundaries of the integral with respect to z is the segment [0, 4] Since the other plane is y = 5 by the similar reasoning we deduce 𝑦 ∈ [0,5]. Finally, we integrate: 5 4 4 640 ∫ ∫ 16 − 𝑥 2 𝑑𝑥𝑑𝑦 = 5 ∫ 16 − 𝑥 2 𝑑𝑥 = 3 0 0 0 Result: 640 3 Question #14 Let S be the parallelogram determined by the vectors  −2 b1 =   3 and  −2 b2 =   5 and let  6 −3 A=   −3 2  Compute the area of the image of S under the mapping 𝑥 ↦ 𝐴𝑥 Answer: Setup the matrix which spans the parallelogram.  −2 −2 S = [b1b2 ] =   3 5 Apply theorem 9 to obtain area of parallelogram which is |det(S)|, then apply theorem 10 to get the area of T(S) | det ( A) | *| det ( S ) |= 6 −3 −2 −2 * =|12 − 9 | *| −10 + 6 |= 3*4 = 12 −3 2 3 5 Results: 12 Question #15 The set 𝐵 = {1 + 𝑡 2 , 𝑡 + 𝑡 2 , 1 + 2𝑡 + 𝑡 2 } is a basis for 𝑃𝟚 . Find the coordinate vector of 𝑝(𝑡) = 1 + 4𝑡 + 7𝑡 2 relative to B. Answer: Given basis can be written in matrix form (first row - second power, second row - first power, third row free coefficient): 1 1 1  B = 0 1 2 1 0 1  Solve Bx = p(t) using augmented matrix: 1 1 1| 7  0 1 2 | 4    1 0 1|1  , multiply first row with -1 and add to third 1 1 1| 7  ~ 0 1 2 | 4  0 −1 0 | −6  −𝑥2 = −6 ⇒ 𝑥2 = 6 𝑥2 + 2𝑥3 = 4 6 + 2𝑥3 = 4 2𝑥3 = 4 − 6 = −2 𝑥3 = −1 𝑥1 + 𝑥2 + 𝑥3 = 7 𝑥1 + 6 − 1 = 7 𝑥1 = 7 − 5 = 2 2 pB =  6   −1 Results: 2 pB =  6   −1 Question #16 5 5 5    Without calculation, find one eigenvalue and two linearly independent eigenvectors of A = 5 5 5 .   5 5 5 Justify your answer. Answer: The matrix 5 5 5  A = 5 5 5 5 5 5 is not invertible because its columns are linearly dependent. So the number 0 is an eigenvalue of A. Eigenvectors for the eigenvalue 0 are solutions of Ax = 0 and therefore have entries that produce a linear dependence relation among the columns of A. Any nonzero vector (in 𝑅3 ) whose entries sum to 0 will work. Find any two such vectors that are not multiples; for instance, (1, 1, -2), and (1, -1, 0). Question #17 Find the absolute maximum and minimum values of 𝑓(𝑥, 𝑦) = 4𝑥𝑦 2 − 𝑥 2 𝑦 2 − 𝑥𝑦 3 Answer: See my photo solution: Question #18 The curvature given by the vector function r is |𝑟 ′ (𝑡) × 𝑟 ′′ (𝑡)| 𝑘(𝑡) = |𝑟 ′ (𝑡)|3 Use the formula to find the curvature of 𝑟(𝑡) =≪ √15𝑡, 𝑒 𝑡 , 𝑒 −𝑡 ≫ at the point (0, 1, 1) Answer: Formula for curvature is 𝑘(𝑡) = ⃗⃗⃗⃗′(𝑡)×𝑟 ⃗⃗⃗⃗⃗⃗′′ (𝑡)|| ||𝑟 3 ⃗⃗⃗⃗′(𝑡)||) (||𝑟 Find first and second derivatives. 1 First derivative is ⃗⃗⃗ 𝑟 ′ (𝑡) = (√15, 𝑒 𝑡 , − 𝑡) 𝑒 Second derivative is 𝑟⃗⃗⃗⃗′′ (𝑡) = (0, 𝑒 𝑡 , 𝑒 −𝑡 ) 2 2 1 Now, find norm (length) of ⃗⃗⃗ 𝑟 ′ (𝑡): ||𝑟⃗⃗⃗′ (𝑡)|| = √(√15) + (𝑒 𝑡 )2 + (− 𝑡) = √2 cos ℎ (2𝑡) + 15 𝑒 Thus, (||𝑟⃗⃗⃗′ (𝑡)||) 3 = (2 cos ℎ (2𝑡) + 15) 3 2 Next, find cross product of first and second derivatives: vec r (t )  r (t ) = (2, − 15 , 15et et Now, find norm (lenght) of 𝑟⃗⃗⃗′ (𝑡) × ⃗⃗⃗⃗ 𝑟 ′′ (𝑡): 2 ||𝑟⃗⃗⃗′ (𝑡) × ⃗⃗⃗⃗ 𝑟 ′′ (𝑡)|| = √(2)2 + (− Finally, curvature is 𝑘(𝑡) = 2 √15 ) + (√15𝑒 𝑡 ) = √30 cos ℎ (2𝑡) + 4 𝑡 𝑒 √30 cos ℎ(2𝑡)+4 15 cos ℎ(2𝑡)+2 3 (√2𝑐𝑜𝑠ℎ(2𝑡)+15) = √2√(2 cos ℎ(2𝑡)+15)3 Now, find curvature at specific point √2 𝑘(0) = 17 𝑘(𝑡) = √2√ 𝑘(0) = 15 cos ℎ (2𝑡) + 2 , (2 cos ℎ (2𝑡) + 15)3 √2 ≈ 0.0831890330807703 17 Question #19 The work W done by a constant force F in moving an object from a point A in space to a point B in space is defined as 𝑊 = 𝐹 ⋅ ⃗⃗⃗⃗⃗ 𝐴𝐵 . Find the work done by a force of 3 newtons acting in the direction 2i+j+2k in moving an object 2 meters from (0,0,0) to (0,2,0) . Answer: Step 1 First we find F, which can be obtained by multiplying its magnitude with the unit vector of the direction it's acting upon. F = 3  2,1, 2   2,1, 2   2,1, 2  = 3 = 3 = 2,1, 2  2 2 2 || 2,1, 2 || 3 2 +1 + 2 Step 2 Next we find ⃗⃗⃗⃗⃗ 𝐴𝐵 , which can be obtained by subtracting the from'' coordinates from the to'' coordinates. ⃗⃗⃗⃗⃗ 𝐴𝐵 = (0,2,0) − (0,0,0) =≪ 0,2,0 ≫ Step 3 Finally we can use the given formula 𝑊 = 𝐹 ⋅ ⃗⃗⃗⃗⃗ 𝐴𝐵 to find the work done. 𝑊 = ≪ 2, 1, 2 ≫⋅≪ 0,2,0 ≫ = 0 + 2 + 0 = 2 𝑗𝑜𝑢𝑙𝑒𝑠 Question #20 Use the cross product to find the sine of the angle between the vectors u=(2,3,-6) and v =(2,3,6)). Answer: From Formula (6) we have: ||𝑢 × 𝑣|| = ||𝑢||||𝑣|| sin 𝜃 where 𝜃 ∈ [0, 𝜋]𝜃 ∈ [0, 𝜋] is the angle between the vectors u and v. This implies sin 𝜃 = ||𝑢×𝑣|| ||𝑢||||𝑣|| In our case we have: u  v = (2,3, −6)  (2,3, 6) = (| 3 −6 2 −6 2 3 |, − | |,| |) = (36, −24, 0) 3 6 2 6 2 3 ||𝑢 × 𝑣|| = ||(36, −24,0)|| = √362 + (−24)2 + 0 = √1872 = 12√13 ||𝑢|| = ||(2,3, −6)|| = √22 + 32 + (−6)2 = √4 + 9 + 36 = √49 = 7 ||𝑣|| = ||(2,3,6)|| = √22 + 32 + 62 = √4 + 9 + 36 = √49 = 7 Thus sin 𝜃 = 12√13 49 ≈ 0.8830