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Using the formula, Area = " y(t)x'(t)dt, find the area under the following parametric curve. x = 2 sin(t),y = 3 cos²(t) for 0 sts

Using the formula, Area = " y(t)x'(t)dt, find the area under the following parametric curve. x = 2 sin(t),y = 3 cos²(t) for 0 sts

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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**Finding the Area Under a Parametric Curve** To find the area under a parametric curve using the given formula, follow these steps: **Formula:** \[ \text{Area} = \int_{a}^{b} y(t) x'(t) \, dt \] Given the parametric equations: \[ x = 2 \sin(t), \ y = 3 \cos^2(t) \] And the interval: \[ 0 \leq t \leq \frac{\pi}{2} \] 1. **Determine \( x'(t) \):** \[ x = 2 \sin(t) \] Differentiating with respect to \( t \): \[ x'(t) = 2 \cos(t) \] 2. **Substitute \( y(t) \) and \( x'(t) \) into the formula:** \[ y(t) = 3 \cos^2(t) \] \[ x'(t) = 2 \cos(t) \] 3. **Set up the integral:** \[ \text{Area} = \int_{0}^{\frac{\pi}{2}} 3 \cos^2(t) \cdot 2 \cos(t) \, dt \] \[ = 6 \int_{0}^{\frac{\pi}{2}} \cos^3(t) \, dt \] 4. **Integrate \( 6 \cos^3(t) \):** Utilize the trigonometric identity for integration: \[ \cos^3(t) = \cos(t) \left( 1 - \sin^2(t) \right) \] Let \( u = \sin(t) \), thus \( du = \cos(t) \, dt \). Therefore, the integral simplifies to: \[ \int_{0}^{\frac{\pi}{2}} \cos^3(t) \, dt = \int_{0}^{1} (1 - u^2) \, du \] \[ = \int_{0}^{1} 1 \, du - \int_{0}^{1} u^2 \, du \] \[ = \left[ u \right]_{0}^{1} - \left[ \frac{u^3}{3} \right]_{0}^{1} \] \[ = (1 -
Transcribed Image Text:**Finding the Area Under a Parametric Curve** To find the area under a parametric curve using the given formula, follow these steps: **Formula:** \[ \text{Area} = \int_{a}^{b} y(t) x'(t) \, dt \] Given the parametric equations: \[ x = 2 \sin(t), \ y = 3 \cos^2(t) \] And the interval: \[ 0 \leq t \leq \frac{\pi}{2} \] 1. **Determine \( x'(t) \):** \[ x = 2 \sin(t) \] Differentiating with respect to \( t \): \[ x'(t) = 2 \cos(t) \] 2. **Substitute \( y(t) \) and \( x'(t) \) into the formula:** \[ y(t) = 3 \cos^2(t) \] \[ x'(t) = 2 \cos(t) \] 3. **Set up the integral:** \[ \text{Area} = \int_{0}^{\frac{\pi}{2}} 3 \cos^2(t) \cdot 2 \cos(t) \, dt \] \[ = 6 \int_{0}^{\frac{\pi}{2}} \cos^3(t) \, dt \] 4. **Integrate \( 6 \cos^3(t) \):** Utilize the trigonometric identity for integration: \[ \cos^3(t) = \cos(t) \left( 1 - \sin^2(t) \right) \] Let \( u = \sin(t) \), thus \( du = \cos(t) \, dt \). Therefore, the integral simplifies to: \[ \int_{0}^{\frac{\pi}{2}} \cos^3(t) \, dt = \int_{0}^{1} (1 - u^2) \, du \] \[ = \int_{0}^{1} 1 \, du - \int_{0}^{1} u^2 \, du \] \[ = \left[ u \right]_{0}^{1} - \left[ \frac{u^3}{3} \right]_{0}^{1} \] \[ = (1 -
### Parametric Curve with Horizontal Tangents #### Problem Statement At what point(s), \((x, y)\), does the following parametric curve have a horizontal tangent? \[ x = 5t + 4, \quad y = 3t^3 - 36t \] #### Explanation In parametric equations, a horizontal tangent occurs when the derivative of \( y \) with respect to \( t \) (denoted \(\frac{dy}{dt}\)) is zero. Given the parametric equations: \[ x = 5t + 4 \] \[ y = 3t^3 - 36t \] First, compute the derivative of \( y \) with respect to \( t \): \[ \frac{dy}{dt} = \frac{d}{dt}(3t^3 - 36t) \] \[ \frac{dy}{dt} = 9t^2 - 36 \] To find where the tangent is horizontal, set \(\frac{dy}{dt} = 0\): \[ 9t^2 - 36 = 0 \] \[ 9t^2 = 36 \] \[ t^2 = 4 \] \[ t = \pm 2 \] Substitute \( t = 2 \) and \( t = -2 \) back into the original parametric equations to find the points \((x, y)\). For \( t = 2 \): \[ x = 5(2) + 4 = 10 + 4 = 14 \] \[ y = 3(2)^3 - 36(2) = 3(8) - 72 = 24 - 72 = -48 \] So, one point is \((14, -48)\). For \( t = -2 \): \[ x = 5(-2) + 4 = -10 + 4 = -6 \] \[ y = 3(-2)^3 - 36(-2) = 3(-8) + 72 = -24 + 72 = 48 \] So, another point is \((-6, 48)\). Therefore, the points where the parametric curve has a horizontal tangent are \( (14, -48) \) and \( (-6, 48) \).
Transcribed Image Text:### Parametric Curve with Horizontal Tangents #### Problem Statement At what point(s), \((x, y)\), does the following parametric curve have a horizontal tangent? \[ x = 5t + 4, \quad y = 3t^3 - 36t \] #### Explanation In parametric equations, a horizontal tangent occurs when the derivative of \( y \) with respect to \( t \) (denoted \(\frac{dy}{dt}\)) is zero. Given the parametric equations: \[ x = 5t + 4 \] \[ y = 3t^3 - 36t \] First, compute the derivative of \( y \) with respect to \( t \): \[ \frac{dy}{dt} = \frac{d}{dt}(3t^3 - 36t) \] \[ \frac{dy}{dt} = 9t^2 - 36 \] To find where the tangent is horizontal, set \(\frac{dy}{dt} = 0\): \[ 9t^2 - 36 = 0 \] \[ 9t^2 = 36 \] \[ t^2 = 4 \] \[ t = \pm 2 \] Substitute \( t = 2 \) and \( t = -2 \) back into the original parametric equations to find the points \((x, y)\). For \( t = 2 \): \[ x = 5(2) + 4 = 10 + 4 = 14 \] \[ y = 3(2)^3 - 36(2) = 3(8) - 72 = 24 - 72 = -48 \] So, one point is \((14, -48)\). For \( t = -2 \): \[ x = 5(-2) + 4 = -10 + 4 = -6 \] \[ y = 3(-2)^3 - 36(-2) = 3(-8) + 72 = -24 + 72 = 48 \] So, another point is \((-6, 48)\). Therefore, the points where the parametric curve has a horizontal tangent are \( (14, -48) \) and \( (-6, 48) \).
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