Find the exact length of the curve.
x = 5cost - cos5t, y = 5sint - sin5t,
0 < t < pi
Good old Pythagoras's theorem says that:
ds^2 = dx^2 + dy^2
We can write:
dx = [-5 sin(t) + 5 sin(5t)] dt
dy = [5 cos(t) - 5 cos(t)] dt
ds^2 = dx^2 + dy^2 =
(using sin^2(x) + cos^2(x) = 1 twice for x = t and x = 5t)
25 [2 - 2 sin(t) sin(5t) - 2 cos(t) cos(5t)] dt^2=
50 [1 - cos(5t-t)] dt^2=
50 [1 - cos(4t] dt^2
Then using cos(2x) = 1-2 sin^2(x), this becomes:
ds^2 = 100 sin^2(2t) dt^2 ----->
ds = 10 |sin(2t)| dt
The integral from 0 to pi is twice the integral from 0 to pi/2, which is an interval where sin(2t) is positive:
s = 20 Integral from 0 to pi/2 of sin(2t) dt =
-10 [cos(pi) - cos(0)] = 20
Well, finding the exact length of a curve can be quite entertaining, just like watching a clown perform! Now, let's get down to business.
To find the length of the curve described by the parametric equations x = 5cost - cos5t and y = 5sint - sin5t, we can use the arc length formula. This formula tells us that the length of a curve from point A to point B is given by the integral of the square root of the derivative of x squared plus the derivative of y squared, integrated with respect to t, from the starting value of t to the ending value of t.
So, let's calculate it step by step:
First, let's find the derivative of x and y with respect to t:
dx/dt = -5sint + 5sin5t
dy/dt = 5cost - 5cos5t
Next, we square these derivatives:
(dx/dt)^2 = (-5sint + 5sin5t)^2
(dy/dt)^2 = (5cost - 5cos5t)^2
Now, let's sum these up and take the square root:
√((dx/dt)^2 + (dy/dt)^2) = √((-5sint + 5sin5t)^2 + (5cost - 5cos5t)^2)
Finally, we integrate this expression with respect to t, from 0 to π, to get the exact length of the curve. However, this integral looks a bit complicated and doesn't lend itself to a simple closed-form solution. So, we'll need to resort to numerical methods or computing tools to approximate the length.
Well, I hope you enjoyed this journey into the world of arc lengths! Just remember, math can always use a little bit of clowning around!
To find the exact length of the curve defined by the parametric equations x = 5cost - cos5t and y = 5sint - sin5t, where 0 < t < pi, we can use the arc length formula.
The arc length formula is given by:
L = ∫ sqrt(dx/dt)^2 + (dy/dt)^2 dt
Let's find dx/dt and dy/dt first:
dx/dt = -5sint + 5sin5t
dy/dt = 5cost - 5cos5t
Now, we can substitute these expressions in the arc length formula:
L = ∫ sqrt((-5sint + 5sin5t)^2 + (5cost - 5cos5t)^2) dt
Next, we need to simplify the integrand:
L = ∫ sqrt(25sint^2 - 10sintsin5t + 25sin^2(5t) + 25cost^2 - 10costcos5t + 25cos^2(5t)) dt
L = ∫ sqrt(50 - 10sintsin5t - 10costcos5t) dt
To evaluate this integral, we need to simplify the expression inside the square root. We can use the identity:
sin(A)sin(B) + cos(A)cos(B) = cos(A - B)
Using this identity, we can rewrite the expression inside the square root:
L = ∫ sqrt(50 - 10[sin(5t)sin(t) + cos(5t)cos(t)]) dt
L = ∫ sqrt(50 - 10cos(5t - t)) dt
L = ∫ sqrt(50 - 10cos(4t)) dt
Now we can integrate.
L = ∫ sqrt(50 - 10cos(4t)) dt
To solve this integral, we can use a trigonometric substitution. Let's substitute u = 4t, then du = 4 dt.
L = (1/4) ∫ sqrt(50 - 10cos(u)) du
L = (1/4) ∫ sqrt(50 - 10cos(u)) du
L = (1/4) ∫ sqrt(50 - 10cos(u)) du
L = (1/4) ∫ sqrt(10(5 - cos(u))) du
L = (1/4) ∫ sqrt(10(sin^2(u/2) + cos^2(u/2) - 2sin^2(u/2))) du
L = (1/4) ∫ sqrt(10(1 - sin^2(u/2))) du
L = (1/4) ∫ sqrt(10cos^2(u/2)) du
L = (1/4) ∫ √10 * cos(u/2) du
Integrating this expression gives:
L = (1/4) * (√10) * ∫ cos(u/2) du
L = (1/4) * (√10) * 2 sin(u/2) + C
L = (√10/2) sin(u/2) + C
Substituting back u = 4t, we have:
L = (√10/2) sin(2t) + C
Now, we can evaluate the limits of integration from 0 to π:
L = (√10/2) [sin(2π) - sin(0)]
L = (√10/2) [0 - 0]
L = 0
Therefore, the exact length of the curve between 0 < t < π is 0.
To find the exact length of the curve given by the parametric equations x = 5cost - cos5t and y = 5sint - sin5t, where 0 < t < pi, we can use the arc length formula.
The arc length formula for a parametric curve is given by:
L = ∫sqrt(dx/dt)^2 + (dy/dt)^2 dt (where ∫ represents integration)
So, let's find dx/dt and dy/dt first.
dx/dt = -5sint + 5sin5t
dy/dt = 5cost - 5cos5t
Now, we can substitute these values into the arc length formula:
L = ∫sqrt((-5sint + 5sin5t)^2 + (5cost - 5cos5t)^2) dt
Now we can simplify this equation and integrate it from 0 to pi to find the exact length of the curve. However, it's worth noting that the integral of this expression is quite complex and requires advanced techniques such as numerical integration or special functions to be evaluated.
Alternatively, if you have access to mathematical software like Wolfram Alpha or MATLAB, you can input this integral expression and it will provide the numerical value as an output.
Therefore, the exact length of the curve can be found through numerical integration or by using mathematical software.
ds = sqrt (dx^2 + dy^2)
dx = -5sin t +5 sin5t
dx^2= 25sin^2 t -50sint sin5t+25sin^2 5t
dy = 5 cos t -5 cos 5t
dy^2= 25cos^2 t-50costcos5t +25 cos^2 5t
luckily sin^2 + cos^2 = 1
dx^2+dy^2=50-50(sin t sin 5t+costcos 5t)
but sin a sin b + cos a cos b = cos(a-b)
= 50 - 50 cos 4 t = 50 (1 - cos 4t)
so
50 integral 0 to pi of (1-cos 4t)^.5 dt
yuuk ! better check my arithmetic.