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
Optimization
Optimization comes from the same root as "optimal". "Optimal" means the highest. When you do the optimization process, that is when you are "making it best" to maximize everything and to achieve optimal results, a set of parameters is the base for the selection of the best element for a given system.
Integration
Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integration and differentiation are inverse operators. Differential equations give a relation between a function and its derivative.
Application of Integration
In mathematics, the process of integration is used to compute complex area related problems. With the application of integration, solving area related problems, whether they are a curve, or a curve between lines, can be done easily.
Volume
In mathematics, we describe the term volume as a quantity that can express the total space that an object occupies at any point in time. Usually, volumes can only be calculated for 3-dimensional objects. By 3-dimensional or 3D objects, we mean objects that have length, breadth, and height (or depth).
Area
Area refers to the amount of space a figure encloses and the number of square units that cover a shape. It is two-dimensional and is measured in square units.
![**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 -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F776a07ce-21ba-443e-aa37-00f610ca7adb%2F79ce66f8-fd96-4476-97d8-4a3263aa4a69%2Fds90g6_processed.png&w=3840&q=75)
![### 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 \( (-6, 48) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F776a07ce-21ba-443e-aa37-00f610ca7adb%2F79ce66f8-fd96-4476-97d8-4a3263aa4a69%2F4eycoig_processed.png&w=3840&q=75)
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