Tag: Applications

Find a Cartesian equation for the pursuit path for given parameters

by RoRi

February 23, 2016

Let Q be a point which moves upward along the positive y-axis and let P be a point which starts at (1,0) and pursues Q according to an equation which stipulates that the distance from P to the y-axis is exactly \frac{1}{k} times the distance from Q to the origin for $k$ a positive number. Find the Cartesian equation for the path of pursuit the point P traces out.

Incomplete.

Find a Cartesian equation for the pursuit path of given parameters

by RoRi

February 23, 2016

Let $Q$ be a point which moves upward along the positive $y$ -axis and let $P$ be a point which starts at $(1,0)$ and pursues $Q$ according to an equation which stipulates that the distance from $P$ to the $y$ -axis is exactly $\frac{1}{2}$ the distance from $Q$ to the origin. Find the Cartesian equation for the path of pursuit the point $P$ traces out.

Incomplete.

Find the orthogonal trajectories of the family of all circles passing through (1,1) and (-1,-1)

by RoRi

February 23, 2016

Find the orthogonal trajectories of the family of curves consisting of all circles passing through the points $(1,1)$ and $(-1,-1)$ .

In a previous exercise (section 8.22, Exercise #12) we found that the family of all circles passing through the points $(1,0)$ and $(-1,0)$ satisfy the differential equation

$(x^2 + 2xy - y^2 - 2)y' - y^2 - 2xy + x^2 + 2 = 0 \quad \implies \quad y' = \frac{y^2 + 2xy - x^2 - 2}{x^2 + 2xy - y^2 - 2}.$

Therefore, the orthogonal trajectories satisfy the differential equation

$y' = \frac{y^2 - x^2 - 2xy + 2}{y^2 - x^2 + 2xy - 2}.$

Incomplete.

Find the orthogonal trajectories of the family of all circles passing through the points (1,0) and (-1,0)

by RoRi

February 23, 2016

Find the orthogonal trajectories of the family of curves consisting of all circles passing through the points $(1,0)$ and $(-1,0)$ .

In a previous exercise (section 8.22, Exercise #11) we found that the family of all circles passing through the points $(1,0)$ and $(-1,0)$ satisfy the differential equation

$(x^2 - y^2 -1) y' - 2xy = 0 \quad \implies \quad y' = \frac{2xy}{x^2-y^2-1}.$

Therefore, the orthogonal trajectories satisfy the differential equation

$y' = \frac{2xy}{y^2 - x^2 + 1}.$

Incomplete.

Find the orthogonal trajectories of the family y = C cos x

by RoRi

February 23, 2016

Find the orthogonal trajectories of the family of curves given by

$y = C \cos x.$

First, we find that the family of curves satisfies the differential equation

$y = C \cos x \quad \implies \quad y' = -C \sin x.$

Since $C = \frac{y}{\cos x}$ we then have

$y' = - \frac{y \sin x}{\cos x}.$

Therefore, the orthogonal trajectories satisfy the differential equation

$\begin{align*} y' = \frac{\cos x}{y \sin x} && \implies && yy' &= \cot x \\ && \implies && \int y \, dy &= \int \cot x \, dx \\ && \implies && \frac{1}{2} y^2 &= \log |\sin x | + C \\ && \implies && y^2 - 2 \log |\sin x| &= C \\ && \implies && y^2 - \log (\sin^2 x) &= C. \end{align*}$

Find the orthogonal trajectories of the family x² – y² = C

by RoRi

February 23, 2016

Find the orthogonal trajectories of the family of curves given by

$x^2 - y^2 = C.$

First, we find that the family of curves satisfies the differential equation

$x^2 - y^2 = C \quad \implies \quad 2x - 2yy' = 0 \quad \implies \quad y' = \frac{x}{y}.$

Therefore, the orthogonal trajectories satisfy the differential equation

$\begin{align*} y' = -\frac{y}{x} && \implies && \frac{1}{y} y' &= -\frac{1}{x} \\ && \implies && \int \frac{1}{y} \, dy &= -\int \frac{1}{x} \, dx \\ && \implies && \log |y| &= -\log|x| + C \\ && \implies && y &= \frac{C}{x} \\ && \implies && xy &= C. \end{align*}$

Find the orthogonal trajectories for the family y = Ce^-2x

by RoRi

February 23, 2016

Find the orthogonal trajectories of the family of curves given by

$y = Ce^{-2x}.$

First, this family of curves satisfies the differential equation

$y = Ce^{-2x} \quad \implies \quad y' = -2Ce^{-2x}.$

Since $C = ye^{2x}$ we then have

$y' = -2y.$

Hence, the orthogonal trajectories satisfy the differential equation

$\begin{align*} y' = \frac{1}{2y} && \implies && 2yy' &= 1 \\ && \implies && 2 \int y \, dy & = \int dx \\ && \implies && y^2 &= x + C. \end{align*}$

Find the orthogonal trajectories of the family x²y = C

by RoRi

February 23, 2016

Find the orthogonal trajectories of the family of curves given by

$x^2y = C.$

From the family of curves, we find that all of the curves in the family satisfy the differential equation

$2xy + x^2y' = 0 \quad \implies \quad y' = \frac{-2y}{x}.$

Therefore, the orthogonal trajectories satisfy the differential equation

$y' = \frac{x}{2y}.$

Hence we have

$\begin{align*} y' = \frac{x}{2y} && \implies && 2y y' &= x \\ && \implies && 2 \int y \, dy &= \int x \, dx \\ && \implies && y^2 &= \frac{1}{2}x^2 + C \\ && \implies && 2y^2 - x^2 &= C. \end{align*}$

Find the orthogonal trajectories of the family y² = Cx

by RoRi

February 23, 2016

Find the orthogonal trajectories of the family of curves by

$y^2 = Cx.$

From the family of curves, we find a differential equation the curves all satisfy,

$y^2 = Cx \quad \implies \quad 2yy' = C.$

Since $C = \frac{y^2}{x}$ we then have,

$2yy' = \frac{y^2}{x} \quad \implies \quad y' = \frac{y}{2x}.$

Therefore, the orthogonal trajectories satisfy the differential equation

$\begin{align*} y' = -\frac{2x}{y} && \implies && yy' &= -2x \\ && \implies && \int y \, dy &= -2 \int x \, dx \\ && \implies && \frac{1}{2}y^2 &= -x^2 + C \\ && \implies && 2x^2 + y^2 &= C. \end{align*}$

Find the orthogonal trajectories of the family x² + y² + 2Cy = 1

by RoRi

February 23, 2016

Find the orthogonal trajectories of the family of curves given by

$x^2 + y^2 + 2Cy = 1.$

Incomplete.