Math Problem Statement

find the length of the parametric curve (x=e^-tcos(t), y=e^-tsin(t)) from -pi<t<0

Solution

The length LL of a parametric curve r(t)=(x(t),y(t))\mathbf{r}(t) = (x(t), y(t)) from t=at = a to t=bt = b is given by the formula:

L=ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt

For the given parametric equations: x(t)=etcos(t)x(t) = e^{-t} \cos(t) y(t)=etsin(t)y(t) = e^{-t} \sin(t)

Step 1: Compute the derivatives of x(t)x(t) and y(t)y(t)

Derivative of x(t)x(t):

dxdt=ddt(etcos(t))=etcos(t)etsin(t)\frac{dx}{dt} = \frac{d}{dt}(e^{-t} \cos(t)) = -e^{-t} \cos(t) - e^{-t} \sin(t)

Derivative of y(t)y(t):

dydt=ddt(etsin(t))=etsin(t)+etcos(t)\frac{dy}{dt} = \frac{d}{dt}(e^{-t} \sin(t)) = -e^{-t} \sin(t) + e^{-t} \cos(t)

Step 2: Compute the magnitude of the velocity vector

Now, we calculate the square of the derivatives:

= e^{-2t} (\cos(t) + \sin(t))^2$$ $$\left( \frac{dy}{dt} \right)^2 = \left( -e^{-t} \sin(t) + e^{-t} \cos(t) \right)^2 = e^{-2t} (\cos(t) - \sin(t))^2$$ Now, adding these: $$\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 = e^{-2t} \left[ (\cos(t) + \sin(t))^2 + (\cos(t) - \sin(t))^2 \right]$$ Simplify the expression inside the brackets: $$(\cos(t) + \sin(t))^2 + (\cos(t) - \sin(t))^2 = 2(\cos^2(t) + \sin^2(t)) = 2$$ Thus, the magnitude of the velocity vector is: $$\sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} = \sqrt{2} e^{-t}$$ ### Step 3: Set up the integral for the length Now, the length of the curve from $$t = -\pi$$ to $$t = 0$$ is: $$L = \int_{-\pi}^0 \sqrt{2} e^{-t} \, dt$$ Factor out $$\sqrt{2}$$: $$L = \sqrt{2} \int_{-\pi}^0 e^{-t} \, dt$$ ### Step 4: Evaluate the integral The integral of $$e^{-t}$$ is: $$\int e^{-t} \, dt = -e^{-t}$$ So, we compute: $$L = \sqrt{2} \left[ -e^{-t} \right]_{-\pi}^0 = \sqrt{2} \left( -e^0 + e^{\pi} \right) = \sqrt{2} \left( -1 + e^{\pi} \right)$$ Therefore, the length of the curve is: $$L = \sqrt{2} \left( e^{\pi} - 1 \right)$$ Would you like further details or have any questions? Here are 5 related questions: 1. What is the geometric interpretation of a parametric curve? 2. How would the curve change if the limits of $$t$$ were different? 3. Can you derive the length formula for parametric curves in 3D space? 4. What is the significance of using parametric equations for curves in physics? 5. How can this process be applied to real-world problems? **Tip**: When working with parametric curves, plotting the curve can give a great insight into its shape and properties.

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Math Problem Analysis

Mathematical Concepts

Calculus
Parametric Curves
Curve Length

Formulas

Length of a parametric curve: L = ∫_a^b sqrt((dx/dt)^2 + (dy/dt)^2) dt
Derivatives of parametric functions: dx/dt, dy/dt

Theorems

Pythagorean theorem in parametric form
Basic integral rules

Suitable Grade Level

Undergraduate Calculus