The fact tells us that this line integral should be the same as the second part ( ie zero) However, let's verify that, plus there is a point we need to make here about the parameterization Here is the parameterization for this curve C 3 → r ( t) = ( 1 − t) 1, 1 t − 1, 1 = 1 − 2 t, 1 for 0 ≤ t ≤ 1 Now, using Green's theorem on the line integral gives, ∮ C y 3 d x − x 3 d y = ∬ D − 3 x 2 − 3 y 2 d A ∮ C y 3 d x − x 3 d y = ∬ D − 3 x 2 − 3 y 2 d A where D D is a disk of radius 2 centered at the origin Since D D is a disk it seems like the best way toThe base is the region enclosed by y = x 2 y = x 2 and y = 9 y = 9 Slices perpendicular to the xaxis are right isosceles triangles The intersection of one of these slices and the base is the leg of the triangle 73 The base is the area between y = x y = x and y = x 2 y = x 2
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The line x+y=2 cuts the parabola- Prove that the curves x = y 2 and xy = k cut at right angles if 8k 2 = 1 Hint Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other Example 2 y = x 2 − 2 The only difference with the first graph that I drew (y = x 2) and this one (y = x 2 − 2) is the "minus 2" The "minus 2" means that all the yvalues for the graph need to be moved down by 2 units So we just take our first curve and move it down 2 units Our new curve's vertex is at −2 on the yaxis
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Find the volume and surface area of a solid of revolution that is generated by rotating the area between the yaxis, y = 0, the line y = 1 and the parabola y = x 2 about the yaxis Plot the Area Clearly, the area is the parabola y = x 2 cut in half and topped by the line y = 1M = 68 m = 78 m = 102 m = 112 c What is the equation, in pointslope form, of the line that is perpendicular to the given line and passes through the point (2, 5)?Surface area and surface integrals (Sect 165) I Review Arc length and line integrals I Review Double integral of a scalar function I The area of a surface in space Review Double integral of a scalar function I The double integral of a function f R ⊂ R2 → R on a region R ⊂ R2, which is the volume under the graph of f and above the z = 0 plane, and is given by
Let f (x,y)=1/ (x^2y^2) for (x,y)\neq 0 Determine whether f is integrable over U0 and over \mathbb {R}^2\bar {U};Answer to Evaluate \iint_S \sqrt {4y 1} ds where S is the first octant part of y = x^2 cut out by 2x y z = 1 By signing up, you'll getAssignment 7 (MATH 215, Q1) 1 Find the area of the given surface (a) The part of the cone z = p x2 y2 below the plane z = 3 Solution The surface can be represented by the vector equation
If so, evaluate Let f (x,y) = 1/(x2 y2) for (x,y) = 0 Determine whether f is integrable over U −0 and over R2 −U ˉ;To y= x2 4, whereas for washers the inner and outer sides would both be determined by y= 4 x2 on the top half of the solid and by y= x 2 4 on the bottom half of the solid Since we're using cylindrical shells and the region runs from x= 2 to x= 2, the volume of the solidY= x 2 z cut o by the plane y= 25 Solution Surface lies above the disk x 2 z in the xzplane A(S) = Z Z D p f2 x f z 2dA= Z Z p 4x2 4y2 1da Converting to polar coords get Z 2ˇ 0 Z 5 0 p 4r2 1rdrd = ˇ=8(101 p 101 1) Section 167 2
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Find the slope of the tangent to a parabola y = x 2 at a point on the curve where x = ½ A 0 93 Find the acute angle that the curve y = 1 – 3x 2 cut the xaxisY= x2 left of x= 1 We reverse the order of integration, so that Z 1 0 Z 1 p y p x3 1 dxdy= Z 1 0 Z x2 0 p x3 1 dydx = Z 1 0 x2 p x3 1 dx = 2 9 (x3 1)3=2j1 0 = 2 9 (23=2 1) c) The integral representing the volume bounded by ˆ= 1 cos˚(in spherical coordinates)X162 Line Integrals 1 Line Integral with respect to the arc length Consider a planar curve C given by r(t) = hx(t), y(t)i, a t b Divide the parameter interval a,b into n subintervals t i 1,t i, i = 0, ,n, and thus divide the curve C correspondingly n subcurves of length Ds
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If the Curve Ay X2 = 7 and X3 = Y Cut Orthogonally at (1, 1), Then a is Equal to (A) 1 (B) −6 6 (D) 0 Mathematics Advertisement Remove all ads Advertisement Remove all adsClick here👆to get an answer to your question ️ If the line y √(3)x 3 = 0 cuts the parabola y^2 = x 2 at A and B , the PA PB is equal to If P = √(3), 0) Ex 63, 23 Prove that the curves 𝑥=𝑦2 & 𝑥𝑦=𝑘 cut at right angles if 8𝑘2 = 1We need to show that the curves cut at right angles Two Curve intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other First we Calculate the point of inters
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historySolution Steps y = x2 y = x − 2 Swap sides so that all variable terms are on the left hand side Swap sides so that all variable terms are on the left hand side x2=y x − 2 = y Add 2 to both sides Add 2 to both sides If the curves ay x^2 = 7 and y = x^3 cut each other orthogonally at a point, find a asked in Limit, continuity and differentiability by SumanMandal ( 546k points) the tangent and normal
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Rewrite as x^22xy=0 This is a quadratic equation in variable x Don't be confused, I'm just pointing out that we will temporarily be thinking of y as a constant (a number) We would solve by factoring if we could, but we can't so we'll use the quadratic formula, which says that the solutions to 2x^2 bx c = 0 are x=(bsqrt(b^24ac))/(2a)This is the 3rd video I sent to a graduating Grade 12 when her graduation in was cancelled because of COVID19 Her Mom asked those who knew her or taugCalculus Multivariable Calculus Find the area of the finite part of the paraboloid y = x 2 z 2 cut off by the plane y = 25 Hint Project the surface onto the xzplane more_vert Find the area of the finite part of the paraboloid y = x 2 z 2 cut off by the plane y
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Divide 0 0 by 4 4 Multiply − 1 1 by 0 0 Add − 2 2 and 0 0 Substitute the values of a a, d d, and e e into the vertex form a ( x d) 2 e a ( x d) 2 e Set y y equal to the new right side Use the vertex form, y = a ( x − h) 2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k kHi Mike, y = x 2 2 is a quadratic equation of the form y = ax 2 bx c, let a = 1, b = 0 and c = 2 You can certainly plot the graph by using values of x from 2 to 2 but I want to show you another way I expect that you know the graph of y = x 2 If you compare the functions y = x 2 and y = x 2 2, call them (1) and (2), the difference is that in (2) for each value of x the You have x2 −y2 = (x y)(x −y) So in your case x2 − y2 x −y = (x y)(x − y) x − y = x y Answer link
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If so, evaluate https//mathstackexchangecom/q/For positive values of a and b, the binomial theorem with n = 2 is the geometrically evident fact that a square of side a b can be cut into a square of side a, a square of side b, and two rectangles with sides a and bWith n = 3, the theorem states that a cube of side a b can be cut into a cube of side a, a cube of side b, three a × a × b rectangular boxes, and three a × b × bSo, other point where parabola will cut the straight line is when x is 2, in which case y = x^2 = (2)^2 = 4 So, second point is (2, 4)
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3,304 2 i think you dropped a sign somewhere, but i would go as follows x 2 y 2 = x 2 y 2 0 = x 2 y 2 ( (xy) (xy)) = x 2 xy xy y 2 but as I think Mark was pointing out, all the steps you took working from RHS to LHS are perfectly valid in Here we can clearly see that the quadratic function y = x^{2} does not cut the xaxis But the graph of the quadratic function y = x^{2} touches the xaxis at point C (0,0) Therefore the zero of the quadratic function y = x^{2} is x = 0 Now you may think that y = x^{2} has one zero which is x = 0 and we know that a quadratic function has 2 zerosOver the region D = {(x,y) x2 y2 8} As before, we will find the critical points of f over DThen,we'llrestrictf to the boundary of D and find all extreme values It is in this second step that we will use Lagrange multipliers The region D is a circle of radius 2 p 2
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X 2 ( 1 2 y) x y 2 y − 2 Find one factor of the form x^ {k}m, where x^ {k} divides the monomial with the highest power x^ {2} and m divides the constant factor y^ {2}y2 One such factor is xy1 Factor the polynomial by dividing it by this factorStart your free trial In partnership withGet stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!
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This is always true with real numbers, but not always for imaginary numbers We have ( x y) 2 = ( x y) ( x y) = x y x y = x x y y = x 2 × y 2 (xy)^2= (xy) (xy)=x {\color {#D61F06} {yx}} y=x {\color {#D61F06} {xy}}y=x^2 \times y^2\ _\square (xy)2 = (xy)(xy) = xyxy = xxyy = x2 ×y2 For noncommutative operators under some algebraicAnswer to Find the area of the finite part of the paraboloid y = x^2 z^2 cut off by the plane y = 81 (Hint Project the surface onto theCalculus Calculus Early Transcendentals Find the area of the finite part of the paraboloid y = x 2 z 2 cut off by the plane y = 25 Hint Project the surface onto the xzplane more_vert Find the area of the finite part of the paraboloid y = x 2 z 2 cut off by the plane y = 25 Hint Project the surface onto the xzplane
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Equation of a Straight Line 11 Solve y3x2 = 0 Tiger recognizes that we have here an equation of a straight line Such an equation is usually written y=mxb ("y=mxc" in the UK) "y=mxb" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis In this formulaFind the area of the finite part of the paraboloid y = x2 z2 cut off by the plane y = 16 Hint Project the surface onto the xzplane Expert Answer % (16 ratings) Previous question Next question Get more help from Chegg Solve it with our calculus problem solver and calculatorCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history
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Y = x 2 2 First, we find the critical points on D We begin by finding the partials and setting them equal to zero • fx(x,y)=1y =0 • fy(x,y)=1x =0 The only critical point on D is (1,1) Notice that f(1,1) = 1 Now, we find the extreme points on the boundary We will use the information in our picture to help us From (0 ,0) to (0,2), theRelated » Graph » Number Line » Examples » Our online expert tutors can answer this problem Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!What must be the value of x so that lines c and d are parallel lines cut by transversal p?
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Factor x^2y^2 x2 − y2 x 2 y 2 Since both terms are perfect squares, factor using the difference of squares formula, a2 −b2 = (ab)(a−b) a 2 b 2 = ( a b) ( aFor graph Y = x^2 Kx 2 to cut x axis y coordinate must be zero thus, x^2 Kx 2 = 0 Now for this equation for solution to be finitly 2 , b^2 4ac > 0 here b = K , a = 1 and c = 2 for every value of K , b^2 4ac will be greater than zero for example k = 0 => (0)^2 4(1)(2) = 8 for k = 1 => (1)^2 4(1)(2) = 912 18 81 99 b Two parallel lines are crossed by a transversal What is the value of m?
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Y = 12 x^2, y = x^2 6Sketch the region enclosed by the given curves Decidewhether to integrate with respect to x or y Draw a typical approximatingrectaExample 57 Find the area of the ellipse cut on the plane 2x 3y 6z = 60 by the circular cylinder x 2 = y 2 = 2x Solution ThesurfaceS liesin theplane 2x3y6z = 60soweusethisto calculatedS =Solve Quadratic Equation by Completing The Square 22 Solving x26x10 = 0 by Completing The Square Subtract 10 from both side of the equation x26x = 10 Now the clever bit Take the coefficient of x , which is 6 , divide by two, giving 3 , and finally square it giving 9 Add 9 to both sides of the equation On the right hand side we have
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