Math Problem Statement

  1. Find the area bounded by the curve x = cos t, y = e t , 0 ≤ t ≤ π 2 and the lines y = 1 and x = 0

Solution

To find the area bounded by the given parametric curve x=cost,y=etx = \cos t, y = e^t for 0tπ20 \leq t \leq \frac{\pi}{2}, and the lines y=1y = 1 and x=0x = 0, we need to set up an appropriate integral. We will follow these steps:

  1. Describe the region: The region is bounded by the curve x=cost,y=etx = \cos t, y = e^t, the vertical line x=0x = 0, and the horizontal line y=1y = 1.

  2. Find the parametric expressions for the area: The area under a parametric curve x=f(t),y=g(t)x = f(t), y = g(t) from t=at = a to t=bt = b is given by: A=abg(t)dxdtdtA = \int_{a}^{b} g(t) \frac{dx}{dt} \, dt Here, g(t)=etg(t) = e^t and f(t)=costf(t) = \cos t.

  3. Calculate dxdt\frac{dx}{dt}: dxdt=sint\frac{dx}{dt} = -\sin t

  4. Set up the integral: The area is given by: A=0π2et(sint)dtA = \int_{0}^{\frac{\pi}{2}} e^t (-\sin t) \, dt Simplifying, this becomes: A=0π2etsintdtA = -\int_{0}^{\frac{\pi}{2}} e^t \sin t \, dt

  5. Evaluate the integral: We need to solve 0π2etsintdt\int_{0}^{\frac{\pi}{2}} e^t \sin t \, dt. We can use integration by parts, where we set u=sintu = \sin t and dv=etdtdv = e^t dt, leading to du=costdtdu = \cos t dt and v=etv = e^t.

    Applying integration by parts: etsintdt=etsintetcostdt\int e^t \sin t \, dt = e^t \sin t - \int e^t \cos t \, dt

    Repeating integration by parts for etcostdt\int e^t \cos t \, dt, set u=costu = \cos t and dv=etdtdv = e^t dt, leading to du=sintdtdu = -\sin t dt and v=etv = e^t: etcostdt=etcostet(sint)dt\int e^t \cos t \, dt = e^t \cos t - \int e^t (-\sin t) \, dt This results in: etsintdt=etsint(etcostetsintdt)\int e^t \sin t \, dt = e^t \sin t - (e^t \cos t - \int e^t \sin t \, dt) Combining terms gives: 2etsintdt=et(sintcost)2 \int e^t \sin t \, dt = e^t (\sin t - \cos t) Therefore, etsintdt=et(sintcost)2\int e^t \sin t \, dt = \frac{e^t (\sin t - \cos t)}{2}

  6. Evaluate at the bounds: A=[et(sintcost)2]0π2A = -\left[\frac{e^t (\sin t - \cos t)}{2}\right]_{0}^{\frac{\pi}{2}} Calculating this, we get: A=12[eπ2(sinπ2cosπ2)e0(sin0cos0)]A = -\frac{1}{2} \left[e^{\frac{\pi}{2}} \left(\sin \frac{\pi}{2} - \cos \frac{\pi}{2}\right) - e^0 \left(\sin 0 - \cos 0\right)\right] =12[eπ2(10)1(01)]= -\frac{1}{2} \left[e^{\frac{\pi}{2}} (1 - 0) - 1 (0 - 1)\right] =12(eπ2+1)= -\frac{1}{2} \left(e^{\frac{\pi}{2}} + 1\right) Thus, the area is: A=1+eπ22A = \frac{1 + e^{\frac{\pi}{2}}}{2}

The area bounded by the curve and the given lines is 1+eπ22\frac{1 + e^{\frac{\pi}{2}}}{2}.

Would you like more details or have any questions?

Here are five related questions:

  1. How would you find the area for different bounds of tt?
  2. What if the parametric curve was given as a different function of tt?
  3. How does integration by parts help in solving parametric integrals?
  4. What are other methods to find the area under a parametric curve?
  5. How do you handle cases where the curve intersects the boundary lines?

Tip: For parametric curves, always ensure to correctly set up the integral based on the derivatives of xx and yy.

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Math Problem Analysis

Mathematical Concepts

Parametric equations
Definite integrals
Integration by parts

Formulas

Area under parametric curve: A = ∫ g(t) (dx/dt) dt
Integration by parts: ∫ u dv = uv - ∫ v du

Theorems

Fundamental Theorem of Calculus
Integration by Parts

Suitable Grade Level

Grades 11-12 / Early College