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Is there anything special about the tangents to the curves \(y^{2}=x^{3}\) and \(2 x^{2}+3 y^{2}=5\) at the points \((1, \pm 1) ?\) Give reasons for your answer.

Short Answer

Expert verified
The tangents are perpendicular at each point.

Step by step solution

01

Identify the given curves and points

The two curves given are \( y^2 = x^3 \) and \( 2x^2 + 3y^2 = 5 \). We are interested in the tangents at the points \((1, 1)\) and \((1, -1)\).
02

Compute the derivative of the first curve

Differentiate the equation \( y^2 = x^3 \) implicitly with respect to \(x\):\[ 2y \frac{dy}{dx} = 3x^2 \]Solving for \( \frac{dy}{dx} \), we get:\[ \frac{dy}{dx} = \frac{3x^2}{2y} \]
03

Find the slope of the tangent to the first curve at points

Substitute \((1, 1)\) into the slope of the tangent:\[ \frac{dy}{dx} = \frac{3(1)^2}{2(1)} = \frac{3}{2} \]Substitute \((1, -1)\) into the slope of the tangent:\[ \frac{dy}{dx} = \frac{3(1)^2}{2(-1)} = -\frac{3}{2} \]
04

Compute the derivative of the second curve

Differentiate the equation \( 2x^2 + 3y^2 = 5 \) implicitly with respect to \(x\):\[ 4x + 6y \frac{dy}{dx} = 0 \]Solving for \( \frac{dy}{dx} \), we get:\[ \frac{dy}{dx} = -\frac{2x}{3y} \]
05

Find the slope of the tangent to the second curve at points

Substitute \((1, 1)\) into the slope of the tangent:\[ \frac{dy}{dx} = -\frac{2(1)}{3(1)} = -\frac{2}{3} \]Substitute \((1, -1)\) into the slope of the tangent:\[ \frac{dy}{dx} = -\frac{2(1)}{3(-1)} = \frac{2}{3} \]
06

Compare the slopes of tangents at the given points

The slope of the tangent to the first curve at \((1, 1)\) is \( \frac{3}{2} \), and at \((1, -1)\) is \(-\frac{3}{2} \). The slope of the tangent to the second curve at \((1, 1)\) is \(-\frac{2}{3} \), and at \((1, -1)\) is \(\frac{2}{3} \). The slopes at \((1, 1)\) are reciprocal negatives of those at \((1, -1)\), indicating a perpendicular relationship at each point.

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Key Concepts

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

Tangents to Curves
When talking about tangents to curves, we refer to lines that lightly touch a curve at just one specific point.
Such lines never cross the curve at that point. Rather, they illustrate the direction in which the curve is heading, much like a compass.
Imagine a curve on a graph as a winding road; a tangent at a certain point would be a path lightly grazing along this road at that exact moment.
To find a tangent line to a curve, you generally need to make use of derivatives.
This is because the tangent's slope at that point coincides with the curve's slope, which is what the derivative tells us.
For example, take the curve defined by the equation \( y^2 = x^3 \). By differentiating this implicitly with respect to \( x \), we get the slope at any given point.
To get the exact tangent line at, say, \((1, 1)\), one would plug these coordinates back into the derivative's result.
In practical terms, finding a tangent means you're pressing pause on the curve's movement and freezing its journey at that spot. This gives you the exact direction, represented as a sloped line.
Slopes of Tangents
Slopes of tangents are critical as they express the steepness and the direction of the tangent line intersecting a curve.
This slope tells us whether the tangent line inclines or declines, and how sharply it does so.
In mathematical terms, the slope is often conceptualized through the derivative using implicit differentiation.
Take the example of the curve \( y^2 = x^3 \). When differentiated implicitly, the derivative or slope at any point \( (x, y) \) becomes \( \frac{3x^2}{2y} \).
For points \((1,1)\) and \((1, -1)\), this results in the slopes \( \frac{3}{2} \) and \(-\frac{3}{2} \), respectively.
The slope at a point is a snapshot of a curve’s behavior at that location; even small changes in position can affect it significantly.
Understanding these slopes gives you insight into how curves twist and turn, and allows us to predict how they'll move forward, much like reading the angle of a hill on a map before you hike it.
Each slope magnifies nuances of the curve at the infinitesimal level.
Perpendicular Lines
Perpendicular lines meet or cross each other at right angles, that is, they make an angle of 90 degrees.
In terms of slopes, two lines are perpendicular if the product of their slopes is \(-1\). This property defines their uniquely intersecting relationship.
Within the problem context, notice that at points \((1, 1)\) and \((1, -1)\) the slopes of tangents on both curves, \( y^2 = x^3 \) and \( 2x^2 + 3y^2 = 5 \), follow this rule.
For the first curve, the slopes are \( \frac{3}{2} \) and \(-\frac{3}{2} \). For the second, they are \(-\frac{2}{3} \) and \(\frac{2}{3} \).
Multiply them: \( \frac{3}{2} \times -\frac{2}{3} = -1 \) at one point. This shows they are perpendicular.
Identifying perpendicular lines using slopes is fundamental in geometry and calculus.
It informs how spaces can be divided, and in some problems, helps determine distances and areas. It’s the mathematical lens through which we analyze how objects relate directionally.
When tackling problems involving tangents, understanding which slopes might turn out to be perpendicular can be a powerful tool.

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