Cylindical Coordinates Infinitesimal Volume The volume, " dV ", is the product of its area, " dA " parallel to the xyplane, and its height, "dz" See the explanantion This is the equation of a circle with its centre at the origin Think of the axis as the sides of a triangle with the Hypotenuse being the line from the centre to the point on the circle By using Pythagoras you would end up with the equation given where the 4 is in fact r^2 To obtain the plot points manipulate the equation as below Given" "x^2y^2=r^2" ">" "x^2y^2 =4 Here are the steps Take the derivative of both sides of the equation with respect to x Separate all of the dy/dx terms from the non dy/dx terms Factor out the dy/dx Isolate dy/dx Some of these examples will be using product rule and chain rule to find dy/dx
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X^2+y^2=49 center and radius-) we can substitute in to obtain (rcos(θ) − 2)2 (rsin(θ))2 = 4 ⇒ r2cos2(θ) −4rcos(θ) 4 r2sin2(θ) = 4 ⇒ r2(cos2(θ) sin2(θ)) − 4rcos(θ) = 0SOLUTION 13 Begin with x 2 xy y 2 = 1 Differentiate both sides of the equation, getting D ( x 2 xy y 2) = D ( 1 ) , 2x ( xy' (1)y) 2 y y' = 0 , so that (Now solve for y' ) xy' 2 y y' = 2x y, (Factor out y' ) y' x 2y = 2 x y, and the first derivative as




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It means a function x of x1x*(x1) is rightUse the ^ (caret) for exponentiation x^2 means x squared x/2y means Resources Simplifier Portal, help with entering simplifier formulas (a must read) The number of common tangents of the circles S ≡ x^2 y^2 4 = 0 and S ≡ x^2 y^2 6x 8y 24 = 0 is asked in Mathematics by SudhirMandal ( 536k points) circle 3 In Mathematica tongue x^2 y^2 = 1 is pronounced as x^2 y^2 == 1 x^2y^2=1 It is a hyperbola, WolframAlpha is verry helpfull for first findings, The Documentation Center (hit F1) is helpfull as well, see Function Visualization, Plot3D x^2 y^2 == 1, {x, 5, 5}, {y, 5, 5}
Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyFree Parabola calculator Calculate parabola foci, vertices, axis and directrix stepbystepAs we had gured out last time, the projection was the disk x2 y2 4 We can write an iterated integral in polar coordinates to describe this disk the disk is 0 r 2, 0 < 2ˇ, so our iterated integral will just be Z 2ˇ 0 Z 2 0 (inner integral) r dr d Therefore, our nal answer is Z 2ˇ 0 Z 2 0 Z 8 r2 r2 (rcos )(rsin )zrdzdrd 2
The simultanous equation calculator helps you find the value of unknown varriables of a system of linear, quadratic, or nonlinear equations for 2, 3,4 or 5 unknowns A system of 3 linear equations with 3 unknowns x,y,z is a classic example This solve linear equation solver 3 unknowns helps you solve such systems systematically 25 T Evaluate Green's theorem using a computer algebra system to evaluate the integral \(\displaystyle \int_C xe^y\,dxe^x\,dy\), where \(C\) is the circle given by \(x^2y^2=4\) and is oriented in the counterclockwise direction 26 Hi Zach Since y^2 = x − 2 is a relation (has more than 1 yvalue for each xvalue) and not a function (which has a maximum of 1 yvalue for each xvalue), we need to split it into 2 separate functions and graph them together So the first one will be y 1 = √ (x − 2) and the second one is y 2 = −√ (x − 2)




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Z = X^2 Y^2;Using integration find the area of the triangle formed by positive xaxis and tangent and normal to the circle x 2 y 2 = 4 at (1, 3 ) View solution Area of the region bounded by y = ∣ Example 10 Find the area of the region enclosed between the two circles 𝑥2𝑦2=4 and (𝑥 –2)2𝑦2=4 First we find center and radius of both circles 𝑥^2 𝑦^2 = 4 〖(𝑥−0)〗^2 〖(𝑦−0)〗^2 = 2^2 Thus, Center = (0, 0) Radius = 2 (𝑥−2)^2 𝑦^2 = 4 〖(𝑥−2)〗^2 〖(𝑦−0)〗^2 = 2^2 Thus, Cente




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After completing the square, we can rewrite the equation as 4(x 21)2 (y 5) 16(z 1)2 = 37 This is a hyperboloid of 1 sheet which has been shifted Speci cally, its centralExplore math with our beautiful, free online graphing calculator Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and moreGraph x^2y^2=4 x2 − y2 = 4 x 2 y 2 = 4 Find the standard form of the hyperbola Tap for more steps Divide each term by 4 4 to make the right side equal to one x 2 4 − y 2 4 = 4 4 x 2 4 y 2 4 = 4 4 Simplify each term in the equation in order to set the right side equal to 1 1 The standard form of an ellipse or hyperbola requires




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Calculus Integral with adjustable bounds example Calculus Fundamental Theorem of Calculus Sketch the curves and identify the region bounded by x = 1/2, x = 2, y = logex and y = 2^x Find area of regionSurfaces and Contour Plots Part 4 Graphs of Functions of Two Variables The graph of a function z = f(x,y) is also the graph of an equation in three variables and is therefore a surfaceSince each pair (x,y) in the domain determines a unique value of z, the graph of a function must satisfy the "vertical line test" already familiar from singlevariable calculus




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How to find dy/dx by implicit differentiation given that x^2 4xy y^2 = 4Here's the 4 simple steps we will take in order to find dy/dx from the given equCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyAnswer and Explanation 1 The sketch of the graph of the equation x2y2 = 4 x 2 y 2 = 4 is shown below Notice that the graph is a circle centered at the



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Graph x^2 (y2)^2=4 x2 (y − 2)2 = 4 x 2 ( y 2) 2 = 4 This is the form of a circle Use this form to determine the center and radius of the circle (x−h)2 (y−k)2 = r2 ( x h) 2 ( y k) 2 = r 2 Match the values in this circle to those of the standard form The variable r r represents the radius of the circle, h h representsSolution The circle \({x^2} {y^2} = 4\) has the radius \(2\) and centre at the origin (Figure \(4\)) Figure 4 Since the upper half of the circle is equivalent to \(y = \sqrt {4 – {x^2}},\) the double integral can be written in the following form Question 34 (Choice 1) Find the area of the region bounded by the curves 𝑥^2𝑦^2=4, 𝑦=√3 𝑥 𝑎𝑛𝑑 𝑥 − 𝑎𝑥𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒




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You must use the "*" (star) symbol for all multiplications!x(x1) is WRONG!Fall 13 S Jamshidi 4 x4 y4 z4 =1 If x,y,z are nonzero, then we can consider Therefore, we have the following equations 1 1=2x2 2 1=2y2 3 1=2z2 4 x4 y4 z4 =1 Remember, we can only make this simplification if all the variables are nonzero!Click here👆to get an answer to your question ️ If R = {(x,y)x, y∈ Z, x^2 y^2≤ 4 } is a relation in Z , then domain of R is



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% Find function value everywhere in the domain contour (X,Y,Z, 4 4) % Plot the isoline where the function value is 4 If you know more about your function and can turn it around into a function of only one variable (eg, sineX^2 y^2 = 4 y = sqrt (x^2 4) x y = 2 y = 2 x Here is the graph The point of intersection of the two graphs is (2, 0)Polynomial identities (short multiplication formulas) (x y) 2 = x 2 2xy y 2 (x y) 2 = x 2 2xy y 2 Example 1 If x = 10, y = 5a (10 5a) 2 = 10 2 2·10·5a (5a) 2 = 100 100a 25a 2 Example 2 if x = 10 and y is 4 (10 4) 2 = 10 2 2·10·4 4 2 = 100 80 16 = 36 The opposite is also true 25 a 4a 2 = 5 2




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A function by definition is a relationship (eg between input x and output y) that fulfils two conditions 1 Every x is related to some y, ie for every input value x there is an output value y 2 Every x is related to at most one y, ie no iAnswer to Use polar coordinates to find the volume of the given solid Inside the sphere x^2 y^2 z^2 = 16 and outside the cylinder x^2 y^2 = Hi Dakota, 1 Substitute for x and y in your equation x=r*cosθ and y=r*sinθ 2 You get r 2 cos 2 θ r 2 sin 2 θ = 4 3 Factor out r 2 on the left side > r 2 (cos 2 θ sin 2 θ)=4



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Free PreAlgebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators stepbystepAn ellipsoid is a surface described by an equation of the form x2 a2 y2 b2 z2 c2 = 1 Set x = 0 to see the trace of the ellipsoid in the yz plane To see the traces in the xy and xz planes, set z = 0 and y = 0, respectively Notice that, if a = b, the trace in the xy plane is a circleSurfaces and Contour Plots Part 2 Quadric Surfaces Quadric surfaces are the graphs of quadratic equations in three Cartesian variables in space




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