Solved: Y=x^2, X=y^2 ; About Y=1 | Chegg.com

à 24 avril

Find the solution of the differential equation that satisﬁes the given initial condition.xy' + y = y^2, y(1) = -1tangent plane to (x^2 + x y + y^2) - 1 at (x,y)=(1,2) SymmetricReduction((x^2 + x y + y^2) - 1, {x^2-1, x}) (x^2 + x y + y^2) - 1 > 0; manipulate c in (x^2 + x y + y^2) - 1 = c; Have a question about using Wolfram|Alpha? Contact Pro Premium Expert Support »Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-stepYou can put this solution on YOUR website! Looking at we can see that the equation is in slope-intercept form where the slope is and the y-intercept is Since this tells us that the y-intercept is .Remember the y-intercept is the point where the graph intersects with the y-axis So we have one point Now since the slope is comprised of the "rise" over the "run" this meansExample 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 y-values 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 y-axis.

x^2+xy+y^2=1 - Wolfram|Alpha

Unfortunately, this question, because of the way it is worded, does not make sense. You can't ask if something is a function without first specifying what you want that something to be a function of. You probably wanted to ask something along the= 29/30 pi Consider the small element width dx as shown, being revolved around the line x = -1 The cross sectional area (csa) of the washer of width dx is the csa of the outer circle minus the csa of the inner ie dA = pi (1 + x + dx)^2 - pi (1 + x)^2 simplify the algebra with u = 1+ x so we have dA = pi ((u+dx)^2 - u^2) = pi (2u dx + dx^2) = pi (2 (1+x) dx + dx^2) we can already see that (dAAssuming that the second expression was meant to be the equation \displaystyle{3}{y}=\frac{{3}}{{2}}{x}-{3} then the two equations are simply variations of the same thing, so infinitely manyFree math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

y=2x+1 - Step-by-Step Calculator - Symbolab

x^2 + xy + y^2 = 1 Find the largest value of y which satisfies the above equation and the corresponding value of x (x is real)x (1+y)^(1/2) = -y (1+x)^(1/2) x^2 (1+y) = y^2 (1+x) y^2 (1+x) - y (x^2) - (x^2) = 0. y = (x^2 +/- sqrt(x^4 + 4(1+x)x^2))/(2(1+x)) Note x=1 gives y=(1 +/- 3)/4 = 1 or -1/2, but only y=-1/2 solves the original problem, so we take the negative branch, at least for x > 0. Now x^4 + 4(1+x)x^2 = x^4 + 4x^3 + 4x^2 = x^2 (x+2)^2, so that2x-y=1 Geometric figure: Straight Line Slope = 2 x-intercept = 1/2 = 0.50000 y-intercept = 1/-1 = -1.00000 Rearrange: Rearrange the equation by subtracting what is to the right of the(c) ( y 2 + x y + 1 ) dx + ( x 2 + x y + 1 ) dy = 0.With M = y 2 + x y + 1 and N = x 2 + x y + 1, note that ( N x - M y) / ( x M - y N ) = ( x - y ) / ( x ( y 2 + x y + 1 ) - y ( x 2 + x y + 1 ) ) = ( x - y ) / ( x - y) = 1. Thus, μ = exp ( ∫ d(xy) ) = e xy is an integrating factor. The transformed equation is: ( y 2 + x y + 1 ) e xy dx + ( x 2 + x y + 1 ) e xy dy = 0.How do you find the volume of a solid y=x^2 and x=y^2 about the axis x=-8? Calculus Using Integrals to Find Areas and Volumes Calculating Volume using Integrals. 1 Answer Leland Adriano Alejandro How do you find the volume of the solid with base region bounded by the curve #y=1-x^2# and the...

shell method:

radius = x - -1 ====> x + 1

y = x^2

y^2 =x =====> y = √(x)

area = √(x) - x^2

intersection:

x^2 = √(x) ===> square both sides

x^4 = x

x^4 - x = 0

x * (x^3 - 1) = 0 ====> x = 0 & 1

∫ 2 * π * (x + 1) * (√(x) - x^2) dx

∫ 2 * π * (x√(x) - x^3 + √(x) - x^2) dx

∫ 2 * π * ( x^(3/2) - x^3 + √(x) - x^2 ) dx

2π * ( x^(3/2 + 1) / (3/2 + 1) - (1/4) * x^4 + x^(1/2 + 1) / (1/2 + 1) - (1/3) * x^3 )

2π * ( x^(5/2) /(5/2) - (1/4) * x^4 + x^(3/2) / (3/2)- (1/3) * x^3

2π * ( (2/5) * x^(5/2) - (1/4) * x^4 + (2/3) * x^(3/2) - (1/3) * x^3

2π * ( (2/5) * x^(5/2) - (1/4) * x^4 + (2/3) * x^(3/2) - (1/3) * x^3 ) limit from 0 to 1

2π * ( (2/5) * (1^(5/2) - 0^(5/2)) - (1/4) * (1^4 - 0^4) + (2/3) * (1^(3/2) - 0^(3/2)) - (1/3) * (1^3 - 0^3)

2π * ( (2/5) * (1 - 0) - (1/4) * (1 - 0) + (2/3) * (1 - 0) - (1/3) * (1 - 0)

2π * ( (2/5) * 1 - (1/4) * 1 + (2/3) * 1 - (1/3) * 1 )

2π * ( (2/5) - (1/4) + (2/3) - (1/3) )

2π * ( 29/60 ) = 29π/30

--------------

y = x^2 ====> √(y) = x

x = y^2

x = x

√(y) = y^2 ----> square both sides

y = y^4

y^4 - y = 0

y ( y^2 - 1 ) = 0

y = 0 & +/- 1 -----> only first quadrant

using the Washer Method:

about x = -1

A (y) = π ( outer radius )^2 - π ( inner radius )^2

A (y) = π ( √(y) - -1 )^2 - π ( y^2 - -1 )^2

A (y) = π ( √(y) + 1 )^2 - π ( y^2 + 1 )^2

A (y) = π [ y + 2√(y) + 1 - ( y^4 + 2y^2 + 1) ]

A (y) = π [ y + 2√(y) + 1 - y^4 - 2y^2 - 1 ]

A (y) = π [ y + 2√(y) - y^4 - 2y^2 ]

∫ π [ y + 2√(y) - y^4 - 2y^2 ] dy

. . . . . . . . . . . . . . . . . . . . .... . . .. . . . . . . . . . . . . . . . ... 1

π [ (1/2) y^2 + 2 * y^(1/2 + 1)/(1/2 + 1) - (1/5) y^5 - (2/3) y^3 ]

. . . . . . . . . . . . . . . . . . . . ..... . . . . . . . . . . .. . . . . . . . .0

. . . . . . . . . . . . . . . . . . . . .... . . . . . . . . . . . . . ... 1

π [ (1/2) y^2 + 2 * y^(3/2)/(3/2) - (1/5) y^5 - (2/3) y^3 ]

. . . . . . . . . . . . . . . . . . . . ..... . .. . . . .. . . . . . . . .0

π [ (1/2) ( 1^2 - 0^2 ) + 2 * (2/3) * ( 1^(3/2) - 0^(3/2) ) - (1/5) (1^5 - 0^5) - (2/3) (1^3 - 0^3) ]

π [ (1/2) ( 1 - 0 ) + (4/3) * ( 1 - 0 ) - (1/5) (1 - 0) - (2/3) (1 - 0) ]

π [ (1/2) * 1+ (4/3) * 1 - (1/5) * 1 - (2/3) * 1 ]

π [ (1/2)+ (4/3) - (1/5) - (2/3) ]

(29π/30)

======

free to e-mail if have a question

1 y y 1 2 y y x correct 3 y y 2 x 2 4 y y x 2 5 y y 3 x 3 ...

Find the volume V of the solid obtained by rotating the ...

6.2 Volume by Cross-Sectional Area; Disk and Washer ...

トップ 100+ Y1 X2 Y0 About The X Axis - イメージ有名

Answered: 12. x 3y 12y 9, x = 0 Sal 29-32 Qn 13.… | bartleby

Calculus I - The Mean Value Theorem

$Calculus I - The Mean Value Theorem$

y axis c y 6 d y 2 e x 2 f x 432 Suppose that the circle 2 ...

r - How to paint the region green? - Stack Overflow

Ilectureonline

Find the volume V of the solid obtained by rotating the ...

Solved: Find The Volume V Of The Solid Obtained By Rotatin ...

Ex 8.2, 3 - Find area bounded by: y = x2 + 2, y = x, x=0,3

Solved: 43. Y = 16 - 2x, Y = 6, X = 0, About X-axis 44. Y ...

Liveatvoxpop: Integral Of 2x1x2

RS Aggarwal Solutions for Class 12 Chapter 17 Area of ...

Solved: Find The Volume V Of The Solid Obtained By Rotatin ...

calculus - Sketch the Region of specific Volume $\frac{512 ...

$calculus - Sketch the Region of specific Volume $\frac{512 ...$

Solved: Y=x^2, X=y^2 ; About Y=1 | Chegg.com

x^2+xy+y^2=1 - Wolfram|Alpha