If x = 18cos^3θ and y = 18sin^3θ, find the total length of the curve swept out by the point (x,y) as θ ranges from 0 to 2π.
just plug and chug:
s = ∫[0,2π] √(x'^2 + y'^2) dθ
Using symmetry, we can take just one arc, and
s = 4∫[0,π/2] √((-54cos^2θsinθ)^2 + (54sin^2θcosθ)^2) dθ
this simplifies quite nicely
Well, well, well... looks like we've got a fancy shmancy math question here! To find the total length of the curve swept out, we need to get our hands a little dirty with some calculus.
First, let's find the derivatives of x and y with respect to theta. The derivative of x with respect to θ will be -54cos^2θsinθ, and the derivative of y with respect to θ will be 54sin^2θcosθ. Now, we can use these derivatives to find the arc length integral.
The arc length integral is given by the formula:
L = ∫ √(dx/dθ)^2 + (dy/dθ)^2 dθ
Now, when we plug in our derivatives, we get:
L = ∫ √((-54cos^2θsinθ)^2 + (54sin^2θcosθ)^2) dθ
And if we simplify that mess, we get:
L = ∫ √(2916sin^4θcos^4θ) dθ
But oh no! This doesn't seem like an easy integral to solve. Looks like the joke's on us this time, my friend.
So, unless you want to spend the rest of your day wrestling with this stubborn integral, I suggest we call in the big guns - a graphing calculator or some computer software. They'll be able to crunch those numbers faster than a clown scarf disappearing up a sleeve!
I hope this response brought a smile to your face, even though I couldn't give you a definitive answer. Remember, math can be tricky, but laughter is the best equation in the world!
To find the total length of the curve using parametric equations x = 18cos^3θ and y = 18sin^3θ, we will use the arc length formula.
The arc length formula is given by:
L = ∫(sqrt(dx/dθ)^2 + (dy/dθ)^2) dθ
Let's find dx/dθ and dy/dθ first:
Differentiating x = 18cos^3θ:
dx/dθ = d(18cos^3θ)/dθ
= -54cos^2θsinθ
Differentiating y = 18sin^3θ:
dy/dθ = d(18sin^3θ)/dθ
= 54sin^2θcosθ
Now, substituting the values of dx/dθ and dy/dθ into the arc length formula, we get:
L = ∫(sqrt((-54cos^2θsinθ)^2 + (54sin^2θcosθ)^2)) dθ
L = ∫(sqrt(2916cos^4θsin^2θ + 2916sin^4θcos^2θ)) dθ
Since cos^2θ = 1 - sin^2θ, we can simplify the expression further:
L = ∫(sqrt(2916(1-sin^2θ)sin^2θ + 2916sin^4θ(1 - sin^2θ))) dθ
Expanding and simplifying the expression, we get:
L = ∫(sqrt(2916sin^2θ - 2916sin^4θ + 2916sin^4θ - 2916sin^6θ)) dθ
L = ∫(sqrt(2916sin^2θ - 2916sin^6θ)) dθ
Taking out common factors, we have:
L = ∫(sqrt(2916sin^2θ(1 - sin^4θ))) dθ
L = ∫(54sinθsqrt(1 - sin^4θ)) dθ
To solve this integral, a substitution can be used to simplify the expression. Let u = sin^2θ, then du = 2sinθcosθdθ:
L = 1/2 ∫(54sqrt(1 - u^2)) du
The integral 1/2 ∫(sqrt(1 - u^2)) du represents the arc length of a semicircle with radius 1. Since we are going from 0 to 2π, we need to find the arc length for a full circle:
L = 1/2 ∫(sqrt(1 - u^2)) du * 2
The integral can now be evaluated using trigonometric substitution or by using a known result for the integral of the form ∫(sqrt(1 - u^2)) du.
The integral is equal to π, so the total length of the curve swept out by the point (x, y) as θ ranges from 0 to 2π is:
L = π.
To find the total length of the curve swept out by the point (x, y) as θ ranges from 0 to 2π, we can use the arc length formula. The arc length formula in polar coordinates is given by:
L = ∫√(r^2 + (dr/dθ)^2) dθ
In this case, r is the distance from the origin to the point (x, y). We can calculate this using the Pythagorean theorem:
r = √(x^2 + y^2)
Dr/dθ is the derivative of r with respect to θ. Let's calculate it:
Dr/dθ = d(√(x^2 + y^2))/dθ
= (1/2)(x^2 + y^2)^(-1/2)(2x(dx/dθ) + 2y(dy/dθ))
= (1/2)(x(dx/dθ) + y(dy/dθ))/(√(x^2 + y^2))
Now, substitute x = 18cos^3θ and y = 18sin^3θ:
Dr/dθ = (1/2)(18cos^3θ((dx/dθ)/dx) + 18sin^3θ((dy/dθ)/dy))/(√((18cos^3θ)^2 + (18sin^3θ)^2))
= (9cos^3θ(dx/dθ) + 9sin^3θ(dy/dθ))/(√(324cos^6θ + 324sin^6θ))
= (9cos^3θ(dx/dθ) + 9sin^3θ(dy/dθ))/(18√(cos^6θ + sin^6θ))
= (cos^3θ(dx/dθ) + sin^3θ(dy/dθ))/(2√(cos^6θ + sin^6θ))
Now, calculate dx/dθ and dy/dθ:
dx/dθ = -54cos^2θsinθ
dy/dθ = 54sin^2θcosθ
Substitute these values back into the equation for Dr/dθ:
Dr/dθ = (cos^3θ(-54cos^2θsinθ) + sin^3θ(54sin^2θcosθ))/(2√(cos^6θ + sin^6θ))
= (-54cos^5θsinθ + 54sin^5θcosθ)/(2√(cos^6θ + sin^6θ))
Now, substitute these values of Dr/dθ and r into the arc length formula and integrate from θ = 0 to 2π:
L = ∫(√(r^2 + (dr/dθ)^2)) dθ
= ∫(√((√(x^2 + y^2))^2 + ((-54cos^5θsinθ + 54sin^5θcosθ)/(2√(cos^6θ + sin^6θ)))^2)) dθ
= ∫(√(x^2 + y^2 + ((-54cos^5θsinθ + 54sin^5θcosθ)^2)/(4(cos^6θ + sin^6θ)))) dθ
Unfortunately, this integral is quite complex and cannot be solved analytically. To find the numerical value of the total length of the curve, you can approximate the integral using numerical methods such as numerical integration techniques or software like MATLAB or Wolfram Mathematica.