Find angle between x^2+y^2 = 16 and y^2 = 6x at intersection?
First you find out the point where those two graphs intersect, then you find the slopes of the tangent lines for those two graphs, and finally get the angle between those two lines. x^2 + y^2 - 16 = y^2 - 6x --> x^2 + 6x - 16 = 0 --> (x + 8)(x - 2) = 0 the graphs meet at x = -8 and x = 2. To find slopes, take the first derivative. 2x dx + 2y dy = 0 and 2y dy = 6 dx When x = 2, y = sqrt(12) so dy/dx = -y/x = -sqrt(12)/2 = -sqrt(3) and dy/dx = 6/[2*sqrt(12)] = sqrt(3)/2 If we consider the point (2,sqrt(12)) as if it were the origin, then the first slope has an angle of -- tan^-1[-sqrt(3)] = -60 degrees The second slope would have an angle of tan^-1[sqrt(3)/2] = 40.89 degrees The angle between these two slopes would thus be 40.89 + 60 = 100.89 degrees. Trying to use the point x = -8 will give complex number results. Besides, x = -8 is not in the domain of y^2 = 6x --> y = sqrt(6x), which requires that 6x > 0 --> x > 0. The other point of intersection will be x = 2 and y = -sqrt(12). The results will be similar and the angle between the slopes will be the same. slope 1 would be sqrt(12)/2 = sqrt(3) slope 2 = 6/[2*-sqrt(12)] = -sqrt(3)/2 tan^-1[sqrt(3)] = 60 degrees tan^-1[-sqrt(3)/2] = -40.89 degrees difference in angles is 100.89 degrees.
What is the angle of intersection of the curves [math]x^2=4y[/math] and [math]y^2=4x[/math] at point [math](0,0)[/math]?
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What will be the angle of intersection of two curves touching each other?
The angle of intersection of curves is defined as the angle between the tangents drawn to each of the curves at the point of intersection.For the better understanding see the following picture,Here θ is the point of intersection of two curves.
How do I show that two curves [math]r^2=a^2\cos^2(\theta)[/math] and [math]r=a(1+cos(\theta))[/math] intersect at an angle?
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Find points of intersection of following lines and curves..?
4x - 3y = 15 so 3y = 4x - 15 squaring yields 9y^2 = 16x^2 - 120 x + 225 tripling yields 27 y^2 = 48 x^2 - 360 x + 675 but 27 y^2 = 8x^2 - 45 so 48 x^2 - 360 x + 675 = 8x^2 - 45 40 x^2 - 360 x + 720 = 0 x^2 - 9 x + 18 = 0 (x - 3) (x - 6) = 0 x = 3 or x = 6 Plugging those into 8x^2 - 27y^2 = 45 yields 27 y^2 = 8x^2 - 45; for x = 3 that is 27 y^2 = 72 - 45 = 27, y^2 = 1, y = +/- 1 and for x = 6 that is 27 y^2 = 288 - 45 = 243, y^2 = 9, y = +/- 3. But plugging those x values into 4x - 3y = 15 gives only (x, y) = (3, -1) or (x, y) = (6, 3) -- for each x, only one of the +/- square roots of y^2 yields a true solution. Going in the other direction, 4x - 3y = 15 so 4x = 3y + 15, therefore 16x^2 = 9y^2 + 90 y + 225. But doubling 8x^2 - 27 y^2 = 45 and rearranging yields 16x^2 = 54 y^2 + 90. So 9y^2 + 90 y + 225 = 54 y^2 + 90 y^2 + 10 y + 25 = 6 y^2 + 10 0 = 5 y^2 - 10 y - 15 0 = y^2 - 2y - 3 = (y - 3)(y + 1) So y = 3 or -1. Plugging into 8x^2 - 27 y^2 = 45 yields x^2 = 36 for y = 3 and x^2 = 9 for y = -1, but plugging into 4x - 3y = 15 shows that only (x, y) = (6, 3) and (3, -1) are actual points of intersection. Dan
Find an equation of the line that bisects the acute angle formed by two lines...?
y = √3.x y = 2 Point of intersection: (2/√3, 2) Slope of y=2 is 0 This line makes an angle of 0° with the x-axis Slope of y=√3.x is √3 This line makes an angle of tanֿ¹(√3) = 60° with the x-axis The acute angle between these lines is 60° The bisector will make an angle of 30° with either of these lines (and also with the x-axis) The slope of the bisector will be tan(30°) = 1/√3 y = x/√3 + c It passes through (2/√3, 2) 2 = 2/3 + c c = 4/3 y = x/√3 + 4/3 3y = √3x + 4 √3x - 3y + 4 = 0 The requried equation is √3x - 3y + 4 = 0
Why must the standard curve intersect the origin of the graph ?
The standard curve will pass through zero (origin) usually only after subtracting the appropriate blank. So, to say that it "MUST" intersect the origin is not a scientifically valid statement. It can be made to go through zero, but it doesn't have to, and normally doesn't.
Find the angle between the curve xy=4 and x^2+y^2=8 at their point of intersection in 3rd and quadrant?
First solve the equations xy= 4( a rectangular hyperbola) and [math]x^2+y^2=8 [/math] (a circle). Put y= [math]\frac{4}{x}[/math] in the 2nd equation. Now, [math]x^2+\frac {16}{x^2}=8[/math] . From here you get two set of solutions x =y=2 and x=y=-2 . These 2 equations intersects at two points (2,2) and (-2,-2). (-2,-2) lies in 3rd quadrant. Now find [math]\frac {dy}{dx}[/math] for both equations at the point (-2,-2). For the 1st one, [math]x\frac{dy}{dx}+y=0[/math] so, (let)[math]s_1=[/math][math]\frac {dy}{dx}=-\frac {y}{x}=-1[/math] (for point (-2,-2)).For 2nd one, [math]x+y\frac {dy}{dx}=0[/math] so (let)[math]s_2[/math][math]\frac {dy}{dx}=-\frac {x}{y}=-1[/math] (for(-2,-2)).Now,[math]\tan\theta=|\frac {s_1-s_2 }{1+s_1×s_2}|[/math]Since [math]s_1=s_2[/math] , they cut at 0° angle, actually they touch each other.
Intersection of cylinder and surface question?
I'm trying to complete the following problem: Consider the surface z = xy and the cylinder x2 + y2 = 1. (a) Find the vector function r(t) that describes the intersection of these two surfaces. (b) At what points is the tangent to r(t) horizontal? (c) Find the equations of the lines that are tangent to r(t) at these points. I have parameterized the vector function into: x = cos(t), y = sin(t), z =cos(t)sin(t) But I can't quite figure out where to go from there. Any help?
Show that the parabolas intersect at right angles.?
get the point(s) of intersection... d/(1 - cosθ) = c/(1 + cosθ) d + d cosθ = c - c cosθ then cosθ = (c- d)/(c+d) the value of r = (c+d)/2 then get the slope at that point slope = dy/dx = (r cosθ + sinθ dr/dθ)/(- r sinθ + cosθ dr/dθ) now , dr/dθ will vary depending on the curve... for the first curve .. . . dr/dθ = c sinθ/(1 + cosθ)^2 second curve dr/dθ = -d sinθ/(1 - cosθ)^2 .. . .. . eventually, determine their respective dy/dx's and note that they would be perpendicular...