/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Show that the curves \(2 x ^ { 2... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that the curves \(2 x ^ { 2 } + 3 y ^ { 2 } = 5\) and \(y ^ { 2 } = x ^ { 3 }\) are orthogonal.

Short Answer

Expert verified
The curves are orthogonal because their tangent lines at points of intersection are perpendicular.

Step by step solution

01

Implicit Differentiation of the First Curve

The equation of the first curve is \(2x^2 + 3y^2 = 5\). To find the slope of the tangent line, differentiate both sides with respect to \(x\), taking into account that \(y\) is a function of \(x\). This yields \(4x + 6yy' = 0\). Solving for \(y'\), the derivative is \(y' = -\frac{2x}{3y}\).
02

Implicit Differentiation of the Second Curve

The equation of the second curve is \(y^2 = x^3\). Differentiating both sides with respect to \(x\) gives \(2yy' = 3x^2\). Solving for \(y'\), we get \(y' = \frac{3x^2}{2y}\).
03

Evaluate the Product of Slopes

For the curves to be orthogonal at points of intersection, the product of the slopes should be \(-1\). The slope of the first curve is \(-\frac{2x}{3y}\) and the slope of the second curve is \(\frac{3x^2}{2y}\). Multiply these slopes: \[\left(-\frac{2x}{3y}\right) \left(\frac{3x^2}{2y}\right) = -\frac{2x \cdot 3x^2}{3y \cdot 2y} = -\frac{6x^3}{6y^2} = -\frac{x^3}{y^2}.\]
04

Substitute from the Second Curve

From the second curve \(y^2 = x^3\), substitute \(x^3\) for \(y^2\) in the product of slopes expression. This gives \(-\frac{x^3}{x^3} = -1\).
05

Conclusion

The product of the slopes is \(-1\), meaning the tangent lines at the points where the curves intersect are perpendicular. Therefore, the curves are orthogonal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Implicit Differentiation
Implicit differentiation is a powerful tool used in calculus to find the derivative of a function when it is not given in the standard form, say, \( y = f(x) \). In implicit differentiation, we consider that \( y \) is a function of \( x \) and differentiate both sides of an equation with respect to \( x \).
For example, in the case of the first curve, \( 2x^2 + 3y^2 = 5 \), we differentiate with respect to \( x \) to find the slope of the tangent line. Using chain rule on \(3y^2\), yields \(6yy'\) because \(y\) is a function of \(x\).
This way of approaching differentiation allows us to solve for derivatives when variables are interdependent. It is a cornerstone concept in analyzing implicit relations between variables.
Tangent Lines
Tangent lines are straight lines that touch a curve at just one point, indicative of the slope at that specified point. In calculus, they are used to approximate the curve near that point.
Determining the equation of a tangent line involves calculating the slope at the point of tangency, which is done by differentiating the equation of the curve. For the curves we analyzed, finding the tangent lines helps assess whether the curves are orthogonal.
To form the tangent line equation, we usually use the point-slope form \( y - y_1 = m(x - x_1) \), where \( m \) is the slope found through differentiation. This concept underpins the idea of orthogonality in curves.
Slope of the Tangent Line
The slope of a tangent line at a point on a curve gives us the rate at which \( y \) changes with respect to \( x \).
  • For the first curve, \( 2x^2 + 3y^2 = 5 \), the slope is found to be \( y' = -\frac{2x}{3y} \).
  • For the second curve, \( y^2 = x^3 \), the slope is \( y' = \frac{3x^2}{2y} \).
These slopes help in understanding how steep or flat the curve is at a given point, which is essential for determining the relationship between two curves.
If the product of the slopes of two tangent lines at their intersection is \(-1\), the lines are perpendicular, hence the curves are orthogonal.
Intersection Points
Intersection points are where two or more curves meet or cross each other. These points are significant because they help us understand relationships, like orthogonality. To find intersection points, set the equations of both curves equal to find common solutions for \( x \) and \( y \).
Checking orthogonality involves examining the slopes of the tangent lines at these intersection points. Each point must ensure the product of the slopes remains \(-1\) to satisfy the orthogonality condition.
Intersection analysis delves deeper into the geometry of curves, making it vital in both calculus and analytical geometry.
Calculus
Calculus is the branch of mathematics focusing on change and motion, leveraging derivatives and integrals to solve problems. It is integral (pun intended) to finding solutions to many complex problems in science and engineering.
In the case of orthogonal curves, calculus allows us to determine the nature of intersections through differentiation and analysis of tangent lines.
Calculus not only handles these technical aspects effectively but also enhances our understanding of the geometric and physical interpretations, essential in the study of dynamical and static systems.

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