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Investigation \(\quad\) Find all \(a\) and \(b\) such that the two curves given by \(y_{1}=a x(b-x)\) and \(y_{2}=\frac{x}{x+2}\) intersect at only one point and have a common tangent line and equal curvature at that point. Sketch a graph for each set of values of \(a\) and \(b\).

Short Answer

Expert verified
The values of \(a\) and \(b\) can be found by solving the three equations obtained from the condition of intersection, equal tangent lines and curvatures. Then the graphs can be sketched using these values.

Step by step solution

01

Setup Intersection Equality

From the description, both curves intersect at a single point. This implies that at that point, the y-values from both equations are equal i.e. \(y_{1}=y_{2}\) or \(a x(b-x)=\frac{x}{x+2}\). This gives the first equation.
02

Setup Tangent Line Equality

The second condition says that both curves share a common tangent line at the point of intersection. The slope of the tangent line is the derivative of the function. For \(y_{1}\), \(\frac{dy_{1}}{dx} = ab - 2ax\). For \(y_{2}\), \(\frac{dy_{2}}{dx} = \frac{2}{(2+x)^2}\). So, \(\frac{dy_{1}}{dx}= \frac{dy_{2}}{dx}\) or \(ab - 2ax = \frac{2}{(2+x)^2}\). This gives the second equation.
03

Setup Curvature Equality

The third condition says that both curves have the same curvature at the point of intersection. Curvature is given by the second derivative of the function. For \(y_{1}\), \(\frac{d^2y_{1}}{dx^2} = -2a\). For \(y_{2}\), \(\frac{d^2y_{2}}{dx^2} = \frac{-4}{(2+x)^3}\). So, \(\frac{d^2y_{1}}{dx^2}= \frac{d^2y_{2}}{dx^2}\) or \(-2a = \frac{-4}{(2+x)^3}\). This gives the third equation.
04

Solve Equations

Solving these three equations simultaneously would yield the values of \(a\) and \(b\).
05

Sketching the Graphs

The final step involves sketching the graph for each pair of \(a\) and \(b\) solutions. Use the resulting coordinates from the solution in the equations of the lines to plot the points and draw the curves.

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Key Concepts

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

Intersection of Curves
When we talk about the intersection of curves, we are referring to the point or points where two different equations meet or cross each other. In this exercise, we want to find the values of \(a\) and \(b\) such that the two curves meet at exactly one point. This means that if we have two equations, \(y_1 = ax(b-x)\) and \(y_2 = \frac{x}{x+2}\), they need to yield the same result for some specific value of \(x\), leading to the equation \(ax(b-x) = \frac{x}{x+2}\).

This setup allows us to explore how the constants \(a\) and \(b\) influence where and if the curves intersect each other. Solving this equation helps us determine the value of \(x\) at the intersection, and confirm that it only happens once. This is important as it can affect how a curve behaves and interacts with another geometrically in a real-world scenario.
Tangents and Derivatives
Tangents and derivatives are closely intertwined concepts in calculus. A tangent line is a line that touches a curve at a single point, without crossing it. This touchpoint is where the curve has the same slope as the tangent line. In calculus terms, this slope is the first derivative of the curve's equation.

In our exercise, we need to ensure that the two curves not only intersect but also share a common tangent line at the point of intersection. This means their derivatives, \(\frac{dy_1}{dx}\) and \(\frac{dy_2}{dx}\), must be equal at that point. For \(y_1\), the derivative is \(ab - 2ax\). For \(y_2\), it is \(\frac{2}{(2+x)^2}\).

Equating the two derivatives, \(ab - 2ax = \frac{2}{(2+x)^2}\), allows us to set up another equation that will help solve for \(a\) and \(b\). Understanding this concept helps us analyze how the slopes of curves change at different points, imagining the behavior of curves as they get closer or further away from each other.
Curvature Analysis
Curvature is a measure of how sharply a curve bends at a particular point. It is mathematically represented by the second derivative of a function. In simple terms, higher curvature means the curve bends more sharply. For our two curves to not only intersect and have a common tangent but also share the same curvature at the intersection point, their second derivatives must coincide.

For the curve \(y_1\), the second derivative is \(-2a\), and for \(y_2\), it is \(\frac{-4}{(2+x)^3}\). Setting these equal, \(-2a = \frac{-4}{(2+x)^3}\), creates another equation needed to find \(a\) and \(b\).

This step adds a layer of understanding to how curves behave not just in terms of position and slope but also in the extent of bending at a single point. Curvature analysis is crucial in fields involving motion, such as physics or engineering, where designing curves with specific properties can be critical.

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Most popular questions from this chapter

Use the model for projectile motion, assuming there is no air resistance. The quarterback of a football team releases a pass at a height of 7 feet above the playing field, and the football is caught by a receiver 30 yards directly downfield at a height of 4 feet. The pass is released at an angle of \(35^{\circ}\) with the horizontal. (a) Find the speed of the football when it is released. (b) Find the maximum height of the football. (c) Find the time the receiver has to reach the proper position after the quarterback releases the football.

Use the given acceleration function to find the velocity and position vectors. Then find the position at time \(t=2\) $$ \begin{array}{l} \mathbf{a}(t)=-\cos t \mathbf{i}-\sin t \mathbf{j} \\ \mathbf{v}(0)=\mathbf{j}+\mathbf{k}, \quad \mathbf{r}(0)=\mathbf{i} \end{array} $$

Use the given acceleration function to find the velocity and position vectors. Then find the position at time \(t=2\) $$ \begin{array}{l} \mathbf{a}(t)=2 \mathbf{i}+3 \mathbf{k} \\ \mathbf{v}(0)=4 \mathbf{j}, \quad \mathbf{r}(0)=\mathbf{0} \end{array} $$

Use the model for projectile motion, assuming there is no air resistance. Use a graphing utility to graph the paths of a projectile for the given values of \(\theta\) and \(v_{0} .\) For each case, use the graph to approximate the maximum height and range of the projectile. (Assume that the projectile is launched from ground level.) (a) \(\theta=10^{\circ}, v_{0}=66 \mathrm{ft} / \mathrm{sec}\) (b) \(\theta=10^{\circ}, v_{0}=146 \mathrm{ft} / \mathrm{sec}\) (c) \(\theta=45^{\circ}, v_{0}=66 \mathrm{ft} / \mathrm{sec}\) (d) \(\theta=45^{\circ}, v_{0}=146 \mathrm{ft} / \mathrm{sec}\) (e) \(\theta=60^{\circ}, v_{0}=66 \mathrm{ft} / \mathrm{sec}\) (f) \(\theta=60^{\circ}, v_{0}=146 \mathrm{ft} / \mathrm{sec}\)

Use the model for projectile motion, assuming there is no air resistance. Find the angle at which an object must be thrown to obtain (a) the maximum range and (b) the maximum height.

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