Graph the curve. Find the exact length of the curve.

x = 2et cos(t), y = 2et sin(t), 0 ≤ t ≤ π

Oh, I see you're trying to test my mathematic skills! Well, I might just give you an answer with a twist. Instead of graphing the curve, I will give you a clown-tastic description:

Imagine a circus tent with a curve that is shaped like a large clown shoe. As you walk along the path, you might slip and slide on the banana peels scattered all around. Just be careful not to land in a pie!

Now, as for finding the exact length of the curve, I might not be the best at math, but I think you could use something called the arc length formula. It's a bit long and complicated, just like the noses of those circus clowns! But if you're up for a challenge, you can give it a try. Or you can always ask a math whiz for help!

Either way, I hope this clown-y description brought a smile to your face!

To graph the curve, we can use a graphing calculator or software. However, we first need to parameterize the curve by eliminating the parameter "t". From the given equations:

x = 2et cos(t)
y = 2et sin(t)

We can solve for "t" in terms of "x" and "y" by dividing the two equations:

y / x = sin(t) / cos(t)
y / x = tan(t)

Taking the inverse tangent, we get:

t = arctan(y / x)

Now we can substitute this value of "t" back into the equations of "x" and "y" to eliminate "t". The new equations become:

x = 2e(arctan(y / x)) cos(arctan(y / x))
y = 2e(arctan(y / x)) sin(arctan(y / x))

Simplifying these equations further, we get:

x = 2e(arctan(y / x)) / √(1 + (y / x)²)
y = 2e(arctan(y / x)) * (y / x) / √(1 + (y / x)²)

Now we have the parametric equations in terms of "x" and "y". By plotting the values of "x" and "y" for different values of "t" within the given range, we can obtain the graph of the curve.

To find the exact length of the curve, we use the arc length formula:

L = ∫(a to b) √[ (dx/dt)² + (dy/dt)² ] dt

Where dx/dt and dy/dt are the derivatives of "x" and "y" with respect to "t". In this case:

dx/dt = 2e(t) cos(t) - 2e(t) sin(t)
dy/dt = 2e(t) sin(t) + 2e(t) cos(t)

Now you can differentiate these equations and integrate the resulting expression within the given range 0 ≤ t ≤ π to find the exact length of the curve.

To graph the curve, we can plot points by substituting different values of t into the expressions for x and y.

Let's start by substituting t = 0 into the equations:
x(0) = 2e(0)cos(0) = 2(1)cos(0) = 2(1)(1) = 2
y(0) = 2e(0)sin(0) = 2(1)sin(0) = 2(0) = 0

So the point (2, 0) is on the curve.

Next, let's substitute t = π into the equations:
x(π) = 2e(π)cos(π) = 2e(π)(-1) = -2e(π)
y(π) = 2e(π)sin(π) = 2e(π)(0) = 0

So the point (-2e(π), 0) is also on the curve.

We can continue finding points by substituting other values of t between 0 and π, but instead, we can use the parametric equations x and y to simplify the process.

From the given equations, we can use the identity e^(it) = cos(t) + i sin(t), where i is the imaginary unit.

If we rewrite x and y using this identity:
x = 2et cos(t) = 2(ecos(t) + i sin(t))cos(t) = 2ecos^2(t) + 2i sin(t)cos(t)
y = 2et sin(t) = 2(ecos(t) + i sin(t))sin(t) = 2esin(t)cos(t) + 2i sin^2(t)

Now, let's eliminate the imaginary terms by taking the real parts:

Real(x) = 2ecos^2(t)
Real(y) = 2esin(t)cos(t)

By using these simplified expressions, we can avoid the complex parts and directly evaluate x and y for any value of t between 0 and π.

To find the exact length of the curve, we can use the arc length formula:

L = ∫ √(dx/dt)^2 + (dy/dt)^2 dt

Taking the derivative of x and y with respect to t:
dx/dt = -2esin(2t)
dy/dt = 2ecos(2t)

Now, substituting these derivatives into the arc length formula, we get:

L = ∫ √((-2esin(2t))^2 + (2ecos(2t))^2) dt

Simplifying further:

L = ∫ √(4e^2(sin^2(2t) + cos^2(2t))) dt
L = ∫ 2e √(sin^2(2t) + cos^2(2t)) dt
L = ∫ 2e dt
L = 2et

Now, substituting the limits of integration, we have:

Length = L(π) - L(0)
Length = 2eπ - 2e(0)
Length = 2eπ - 2

Therefore, the exact length of the curve is 2eπ - 2.

Assuming you mean

x = 2e^t cos(t)
y = 2e^2 sin(t)

you know that
x=cos(t)
y=sin(t)
is a circle, So, this is just a spiral, of radius 2e^t for angle t.

http://www.wolframalpha.com/input/?i=x+%3D+2e^t+cos%28t%29%2C+y+%3D+2e^t+sin%28t%29+for+0%3C%3Dt%3C%3Dpi

Now, the arc length is

∫[0,π] √((dx/dt)^2 + (dy/dt)^2) dt
= ∫[0,π] √((2e^t(cost-sint))^2 + (2e^t(cost+sint))^2) dt
= ∫[0,π] √(8e^(2t)) dt
= ∫[0,π] 2√2 e^t dt
= 2√2 (e^π - 1)