The two curves x = y^2 and xy = a^3 cut orthogonally, then a^2 is equal to Sarthaks eConnect Largest Online Education Community The two curves x = y2 and xy = a3 cut orthogonally, then a2 is equal to (A) 1/3 (B) 3 2 (D) 1/2 LoginParabola Equation Derivation In the above equation, "a" is the distance from the origin to the focusY 2 = – 4ax;
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The line x+y=2 cuts the parabola-Find the slope of the tangent to a parabola y = x 2 at a point on the curve where x = Y 2 = c 2 cut at right angles, then (A) a 2 b 2 = 2 c 2 (B) b 2 ±
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A 2330 b c d 25Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreFor 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 >
Answer 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 theLet the parabolas y = x( c x ) and y=x2btouch each other at the point (1, 0), then View Answer If the curves a2x2 b2y2 =1and 25x2 16y2 =1cut each other orthogonally, then a2−b2= View Answer Find the equation of the tangent to the curve x=θsinθ,y=1cosθat θ=4π View AnswerAnswer Volume V of the solid generated by revolving the area cut off by latus rectum (x = a) of the parabola y^2 = 4ax, about its axis, which is x axis, is
36 ‡ 0 1 ‡2 2 ‡ 0 4y2dz d y dx in the order d y dz dx 37 ‡ 0 1 ‡ 0 1x 1x2d y dz dx in the order dz d y dx 38 ‡ 0 4 ‡ 0 16 x2 0 16 x2z2 d y dz dx in the order dx d y dz 3944 Average value Find the following average values 39 The average temperature in the box D =8x, y, zL 0 §x §ln 2, 0 §y §ln 4, 0 §z §ln 8<{x∈Q x>0 and x2 <2}, B= {x∈Q x>0 and x2 ≥2} In this example neither Anor Bhas an extremal element The first exercise below shows that, with regard to extremal elements, every cut isAnswer (1 of 4) What is the length of the chord cut off by y = 2x1 from the circle x^2y^2=2?
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The other one is a xy that contains both x and y (2) G is in separable (3) For each (e;f) 2E E there is(1)^2 4(1)(2) = 932 2 32 e j Torque z x y T Cut Surface τ τ = T ⋅r J τ π τ π max max = ⋅ ⋅ = ⋅ ⋅ ⋅ − 16 16 3 4 4 T D T D D D for solid circular shafts o for hollow shafts e o i j Rectangular Cross Section Torque z x y T Centroid τ Torsional Stress Cut Surface τ 1 2 a b Note a >
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3 If z = f(x) for some function f(), then –z = jf0(x)j–x We will justify rule 1 later The justification is easy as soon as we decide on a mathematical definition of –x, etc Rule 2 follows from rule 1A 2 = 2 c 2 a 2 ±Find the area of the surface G cut from the hemisphere x2y2z2=42, z≥0, by the plane z=1 and z=3 by setting up an integral Method 1 Cylindrical Coordinates Set up the Integral 2 42−r2 rdrdθ r=7 ∫15 θ=0 ∫2π 2 42−r2 rdrdθ r=7 ∫15 θ0 ∫2π (5 pts) (5 pts) (5 pts) The surface area of the hemisphere between z=1 and z=3 is 1(f x) 2f (y) 2 S ∫= −x
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Co, 0 y 2 Noncircular cylinder A solid right (noncircular) cylinder has its base R in the xyplane and is bounded above by the paraboloid z x2X 2 = 9p(9 y) (i) and x 2 = p(y 1) (ii) As, these curves cut each other at right angle, therefore their tangent at point of intersection are perpendicular to each other So, let us first find the point of intersection and slope of tangents to the curves From Eqs (i) and (ii), we get 9p(9 y) = p(y 1) 9(9 y) = y 1If 2 2 2 2 1() x y a b a b ²!
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Graph y=3x2 y = 3x 2 y = 3 x 2 Use the slopeintercept form to find the slope and yintercept Tap for more steps The slopeintercept form is y = m x b y = m x b, where m m is the slope and b b is the yintercept y = m x b y = m x b Find the values of m m and b b using the form y = m x b y = m x b0 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 =>2, true if there are no cut vertices If x is a cut vertex, write G = x G 1 G 2 G k Each singlevertex G i is a noncut vertex Each nontrivial G i has at least two noncut vertices, one of which is nonadjacent to x;
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Show that the two curves x 2 – y 2 = r 2 and xy = c 2 where c, r are constants, cut orthogonally Solution Let (x 1, y 1) be the point of intersection of the two curves I Curve x 2 – y 2 = r 2 Differentiating wrto x II Curve xy = c 2 Differentiating wrto x Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of DifferentialB Cross Section Method 1Find the area of the surface G cut from the hemisphere x2y2z2=42, z≥0, by the plane z=1 and z=3 Solution The surface of the hemisphere is defined by z=16−x2−y2 z x= −2x 216−x2−y2 z y= −2y 216−x2−y2 The surface area of the hemisphere between z=1 and z=3is 1(z x) 2z (y) 2 S ∫ 1 x 16−x2−y2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 y 16−x2−y2 ⎛ ⎜ ⎞ ⎟ 2 S ∫∫ 16−x2−y2 16−x2−y2 x2
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And bounded on the sides by the cylinder x2 y2 x = 0 Solution V = Zˇ=2 ˇ=2 Zcos 0 (1 r2)rdrd = Zˇ=2 ˇ=2 cos2 2 cos4 2 d = 5ˇ 32 5 Find the mass and centre of mass of the plate that occupies the given region with the given density function (a) =Use Stokes' theorem to evaluate ∫ C 2 x y 2 z d x 2 x 2 y z d y (x 2 y 2 = 2 y i e z j − arctan x k with S as a portion of paraboloid z = 4 − x 2 − y 2 z = 4 − x 2 − y 2 cut off by the xyplane oriented counterclockwise 363 T Use a CAS to evaluate2 is at a distance p x2 o y2 from the same point on the y axis Its field in the y direction is then found similarly, accounting for the geometric factor to pick out the y component E 2y = kq 2 x2 o y2 p x o x2 o y2 = kq 2y (x 2y )3/2 (11) We need E 1y =−E 2y for the net vertical field to be zero (resulting in a purely horizontal field
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Y locations (with respect to the moment from each force) from M x W xW n i y i i 1 W W xdW x xdW x OR W x W xThe area of the smaller part of the circle, x 2 y 2 = a 2, cut off by the line, `x = a/sqrt2`, is the area ABCDA It can be observed that the area ABCD isB 2 = 2 c 2 (D) a 2 b 2 = 2 c 2 13 Let P(a sec T, b tan T) and Q(a sec I, b tan I) where T I = S /2, be two points on the hyperbola 2 2 2 2 1 x y a b ±
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X 2 = – 4ay;The expected cut off of XGroup 1/21 is 2735 (Out of 70) and 3035 (Out of 50) of YGroup 1/21 Candidates who clear the cut off will be called for the second phase Phase 2 consists of four stages that one has to clear Four stages of phase 2 are Join dehradun defence academy for best air force coaching in dehradunX 2 = 4ay;
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Severe weather cut short New York's superstarladen concert 'Homecoming concert' in Central Park, with Barry Manilow's set being interrupted by an announcemeShow Solution Okay, since we are looking for the portion of the plane that lies in front of the y z y z plane we are going to need to write the equation of the surface in the form x = g ( y, z) x = g ( y, z) This is easy enough to do x = 1 − y − z x = 1 − y − z Next, we need to determine just what D D is𝑥𝑦=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 intersection of Curve (1) &
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Apart from these two, the equation of a parabola can also be y 2 = 4ax and x 2 = 4ay if the parabola is in the negative quadrants Thus, the four equations of a parabola are given as y 2 = 4ax;2 x yA y x Resultant force Over a body of constant thickness in x and y W n i F z W i 1 W dW Location x, y is the equivalent location of the force W from all W i's over all x &Transcript Ex 63, 23 Prove that the curves 𝑥=𝑦2 &
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Example 9 Find the area of a minor segment of the circle x2 y2 = a2 cut off by the line x = 2 a Solution Solving the equation x2 y2 = a2 and x = 2 a, we obtain their points of intersection which are , 3 2 2 a a and 3, – 2 2 a aSolution Parameterise C by r(t) = (x(t),y(t) = (4 − t2,t), where −3 ≤ t ≤ 2, since −3 ≤ y ≤ 2 F= (y2,x) and dr= (dx,dy) Hence, Z C FCircular sector Integrate f(x, y) = over the smaller sector cut from the disk x2 4 by the rays = 77/6 and = 7/2, Unbounded region Integrate = — — 1)2/3 over the infinite rectangle 2 x <
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And x 2 ±The definite integral is 2int_3^5sqrt(25 x^2)dx There are always multiple ways to approach integration problems, but this is how I solved this one We know that the equation for our circle is x^2 y^2 = 25 This means that for any x value we can determine the two y values above and below that point on the x axis using y^2 = 25 x^2 y = sqrt(25x^2) If we imagine that a lineWe have a question and this went to find the area of the smaller part area of the smaller part of the circle Access corpus was quite equal to a square cut off by the line x equal to a by two circulars access choir plus y squared equal to a square and line is X equal to a by two Okay, so basically this is the circle and a Beirut is somewhere here
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Dr= Z C y2dx xdy = Z 2 −3 t2 dx dt dt− Z 2 −3 (4−t2) dy dt dt = Z 2 −3 −2t3 (4−t2)dt = 245/6 Example 53 Evaluate the line integral, R C(x 2 y2)dx(4xy2)dy, where C is the straight line segment4xy = 48 x 2 or We wish to MAXIMIZE the total VOLUME of the box V = (length) (width) (height) = (x) (x) (y) = x 2 y However, before we differentiate the righthand side, we will write it as a function of x only Substitute for y getting V = x 2 y = 12x (1/4)x 3 Now differentiate this equation, getting V' = 12 (1/4)3x 2 = 12 (3/4)x 2 = (3/4)(16 x 2)A 0 93 Find the acute angle that the curve y = 1 – 3x 2 cut the xaxis
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A figure is bounded by the curves `y=x^21, y=0,x=0,a n dx=1` At what point `(a , b)` should a tangent be drawn to curve `y=x^21` for it to cut off a trapezBy solving the two equations simultaneously, you get the points of on∬ D (e x 2 y 2 x 4 2 x 2 y 2 y 4) For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure) 164
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In this section we will start evaluating double integrals over general regions, ie regions that aren't rectangles We will illustrate how a double integral of a function can be interpreted as the net volume of the solid between the surface given by the function and the xyQ Find the volume cut off from the cylinder $x^2y^2=ax$by the planes $z=0 $and $z=x$ Given Answer$\frac{128a^3}{15}$ My answer$\frac{\pi a^3}{4}$ Working So, first off, we can decipher from the question that $z$will vary from $0$to $x$ $y$will vary from $\sqrt{axx^2}$to $\sqrt{axx^2}$Two tangent are drawn from P (1, 8) to the circle x 2 y 2 − 6 x − 4 y − 1 1 = 0 touching the circle at A and B A circle passes through the point of intersection of circles x 2 y 2 − 2 x − 6 y 6 = 0 and x 2 y 2 2 x − 6 y 6 = 0 and intersects the circumcircle of Δ PAB orthogonally If r is the radius of circle then find r
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Circulation of the eld F = (y 2 z2)i (x 2 z)j (x y2)k around the curve C The boundary of the triangle cut from the plane x y z= 2 by the rst octant, counterclockwise when viewed from above like x167 #4 Sol Note that the surface S enclosed by C has parametric equations as s(x;y) = xi yj (2 x y)k;Air force x and y group resultAir force new vacancy 21Air force bharti 21Air force result 21Air force cut off 21Air force bharti new update 21Air fEquation y = x3 −3x2 −10x cuts the xaxis Cuts xaxis when y = 0 Therefore 0 = x3 −3x2 −10x Thus 0, 5, 2 0 ( 5)( 2) 0 (2 3 10) ⇒ = = = − ⇒ = − = − − x x x x x x x x x Hence the points are (0, 0), (5, 0) and (2, 0) Exercise Find the points where the graphs given by the following equations cut the xaxis 1 y = 3x
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(2) 𝑥=𝑦2 𝑥𝑦=𝑘 Putting 𝑥=𝑦2 in (2Solution Given y = x 2 4 x 6 Rewrite in vertex form by completing the square y = x 2 4 x 6 = (x 2) 2 10 Start y = x 2 2 units to the right y = (x 2) 2 reflection on the x axis y = (x 2) 2 shift 10 units up y = (x 2) 2 10A 5 m x 5 cm is cut from a corner of cm x 30 cm cardboard Find the centroid from the longest side a 1099 m b 1142 m c 1033 m d 1242 m View Answer The area of the second and third quadrant of the circle x 2 y 2 = 36 is revolved about the line x = 4 What is the volume generated?
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The f (x) curve is above the xaxis from x=0 to the point where 2x = x^2, which is x = 2 The area is INTEGRAL (2x x^2) dx = x^2 x^3/3 @x=2 0 to 2 = 4/3 That agrees with what you obtained You want the area under y = mx to be 2/3
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