MATH 2107 Lecture Notes - Lecture 1: Cross Product, Joule, Parallelogram

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11 Oct 2016

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Solutions to homework assignment #1: sketch the curve r = 1 + cos , 0 2 , and nd the area it encloses. Figure 1: the curve r = 1 + cos , 0 2 . =0 (1 + 2 cos + cos2 : find the dot product ~a ~b in the following cases: (a) ~a =< 1, 0, 2 >,~b =< 2, 0, 1 > . Are these vectors orthogonal? (b) ~a =< x2y3 x3y2, x3y1 x1y3, x1y2 x2y1 >,~b =< x1, x2, x3 >, where the xi, yi are any real numbers. Are these vectors orthogonal? (c) ~a is a unit vector having the same direction as ~i +~j and ~b is a vector of magnitude. 2 in the direction of ~i + ~j ~k. Solution: (a) ~a ~b = 2 2 = 0. These vectors are orthogonal. (b) ~a ~b = (x2y3 x3y2)x1 + (x3y1 x1y3)x2 + (x1y2 x2y1)x3 = 0.

Related Questions

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1) A water tank is in the shape of a paraboloic. Its shape is obtained by rotating the parabola y=.25x², between x=0 and x=4 around the y-axis. x and y are in feet

a) Find the volume for the tank

b) The tank is full of water. Find the work required to pump the water out of the tank ( Water weighs 62.5 lbs per cubic foot)

2) consider the function f(x)=x³-6x² +P, where P is a constant. Find the value of P so that the average value of f over the interval (-1,2) is 1.

3) a) sketch the graph of r=cos(2theta), for 0<theta<2pi

b) Find the area enclosed by one loop of the figure in your sketch

4) Consider the region in the first quadrant bounded by y=4x² , the y-axis, and the line y=4

a) Find the area of this region

b) Find the centroid of this region.

I have the answer for the following question. But some of theanswers are not correct.

Please do the following question by yourself and check which arethe wrong answers,please post your answers and compare to it.

Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S

MATH 2107 Lecture Notes - Lecture 1: Cross Product, Joule, Parallelogram

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