Transform polar equation into rectangular?
A. 1 radian = 180/π degrees = 57.296 degrees Therefore, theta = 76.776 degrees ----- B. r = tan(theta) * sec(theta) r = sin(theta) / cos^2(theta) r * cos^2(theta) = sin(theta) Multiply each side by r r^2 * cos(theta)^2 = r * sin(theta) (r * cos(theta))^2 = r * sin(theta) Since, x = r * cos(theta) and y = r * sin(theta), y = x^2 Done!
Convert the polar equations r = 4 and r = 6 / (2*cos[theta] - 3*sin[theta]) to rectangular form?
(1) By squaring both sides: r^2 = 4^2 ==> r^2 = 16. Then, since r^2 = x^2 + y^2, we see that this equation in rectangular form is: x^2 + y^2 = 16. (2) Multiplying both sides by 2cosθ - 3sinθ gives: r(2cosθ - 3sinθ) = 6 ==> 2r*cosθ - 3r*sinθ = 6. Using x = r*cosθ and y = r*sinθ, this becomes: 2x - 3y = 6. I hope this helps!
Transformation of polar equations help?
well, you know that r cos (theta) is the same as x, so: a) is simply x = 4 and that is a line. With b), multiply either side by r, so: r^2 = 6 r sin (theta), then we know that r sin (theta) is the same as y: r^2 = 6y, and we also know r^2 = x^2 + y^2, therefore: x^2 +y^2 = 6y, that is rectangular... So: a) x = 4 b) x^2 +y^2 - 6y = 0 Actually, for b you could always complete the square... x^2 + y^2 - 6y + 9 = 9 x^2 + (y - 3)^2 = 9 This would make a circle
What is the equation for a semicircle?
In an x–y Cartesian coordinate system, the Circle with centre coordinates (a, b) and radius r is the set of all points (x, y) such thatSo, Upper Half circle be,Lower Half circle be,
What is this question wanting me to do?
Its asking you to do both. 1) r=6 is a circle of radius 6. In rectangular coordinates it is x^2+y^2=36 2) straight line with slope k =inverse tan pi/6. y=kx 3) horizontal line y=3
Can someone please help I really dont get this?
It is helpful to remember the transformation from polar coordinates to rectangular coordinates: .. x = r cos(θ) .. y = r sin(θ) 1a. x = r cos(1.34), y = r sin(1.34) Solving the first equation for r gives .. r = x/cos(1.34) Using this in the second equation gives .. y = (x/cos(1.34)) sin(1.34) .. y = x tan(1.34) .. y = 4.256x ... a straight line through the origin with positive slope 1b. Same deal. .. x = (tan(θ) sec(θ)) cos(θ) = tan(θ) .. y = (tan(θ) sec(θ)) sin(θ) = tan(θ)^2 Using the first equation in the second, we get .. y = x^2 ... a parabola with vertex at (0, 0) 2a. modulus = √((real part)^2 + (imaginary part)^2) = √(0^2 + 5^2) = 5 .. argument = arctan((imaginary part)/(real part)) = arctan(5/0) = π/2 2b. In the form m(cos(a)+i sin(a)), m is the modulus and "a" is the argument. .. modulus = √10 .. argument = 5π/7
How do you find the volume of an ellipsoid?
If the ellipsoid is defined via[math]\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1[/math]which is the standard presentation, then the volume is[math]V=\frac{4}{3}\pi a b c[/math]. The parameters [math]a,b,c[/math] are the lengths of the semi-principal axes, and if you need the volume of an ellipsoid given in some other form it is usually helpful to find them first.