Chapter 1: Q33E (page 1)

Use the method in Problem 32 to find the orthogonal trajectories for each of the given families of curves, where k is a parameter.
(a) $2 x^{2} + y^{2} = k$
(b) $y = {kx}^{4}$
(c) $y = e^{kx}$
(d) $y^{2} = kx$
[Hint: First express the family in the form F(x, y) = k.]

Short Answer

Expert verified

$x = c y^{2}$
role="math" $x^{2} + 4 y^{2} = c$
$2 y^{2} \ln y - y^{2} + 2 x^{2} = c$
$2 x^{2} + y^{2} = C$

Step by step solution

(a): Find the orthogonal trajectory of curve 2x2+y2=k

Here, the curve is $2 x^{2} + y^{2} = k$

$\begin{array}{l} \frac{\partial F}{\partial y} (x, y) = 2 y \\ \frac{\partial F}{\partial x} (x, y) = - 4 x \\ 2 y dx - 4 x dy = 0 \\ \int \frac{1}{x} dx = 2 \int \frac{1}{y} dy = 0 \\ \ln |x| = 2 \ln |y| + c \\ x = c y^{2} \end{array}$

Hence the solution is $x = c y^{2}$

(b): Determine the orthogonal trajectory of curve y=kx4.

Here the curve is $y = k x^{4}$

$\begin{array}{l} \frac{\partial F}{\partial y} (x, y) = \frac{1}{x^{4}} \\ \frac{\partial F}{\partial x} (x, y) = - \frac{4 y}{x^{5}} \\ \frac{1}{x^{4}} dx + \frac{4 y}{x^{5}} dy = 0 \\ \int x dx = - 4 \int y dy = 0 \\ \frac{x^{2}}{2} = - 2 y^{2} + c \\ x^{2} + 4 y^{2} = c \end{array}$

Hence the solution is localid="1664254475026" $x^{2} + 4 y^{2} = c$

(c): Evaluate the orthogonal trajectory of curve y=ekx

Here the curve is $y = k e^{x}$

$\begin{array}{l} \frac{lny}{x} = k \\ \frac{\partial F}{\partial y} (x, y) = \frac{1}{xy} \\ \frac{\partial F}{\partial x} (x, y) = - \frac{\ln y}{x^{2}} \\ \frac{1}{xy} dx + \frac{\ln y}{x^{2}} dy = 0 \\ \int x dx = - \int y \ln y dy = 0 \\ \frac{x^{2}}{2} = \frac{- y^{2} \ln y}{2} + \frac{y^{2}}{4} + c \\ \frac{x^{2}}{2} + \frac{y^{2} \ln y}{2} - \frac{y^{2}}{4} + c \\ 2 x^{2} + 2 y^{2} \ln y - y^{2} = c \\ 2 y^{2} \ln y - y^{2} + 2 x^{2} = c \end{array}$

Hence the solution is $2 y^{2} \ln y - y^{2} + 2 x^{2} = c$

(d): Find the orthogonal trajectory of curve y2=kx

Here the curve is $y^{2} = k x$ .

$\begin{array}{l} \frac{y^{2}}{x} = k \\ \frac{\partial F}{\partial y} (x, y) = \frac{2 y}{x} \\ \frac{\partial F}{\partial x} (x, y) = - \frac{y^{2}}{x^{2}} \\ \frac{2 y}{x} dx + \frac{y^{2}}{x^{2}} dy = 0 \\ 2 \int xdx = - \int ydy = 0 \\ \frac{2 x^{2}}{2} = - \frac{y^{2}}{2} + c \\ 2 x^{2} + y^{2} = c \end{array}$

Hence the solution is $2 x^{2} + y^{2} = c$

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Use the method in Problem 32 to find the orthogonal trajectories for each of the given families of curves, where k is a parameter.
(a) $2 x^{2} + y^{2} = k$
(b) $y = {kx}^{4}$
(c) $y = e^{kx}$
(d) $y^{2} = kx$
[Hint: First express the family in the form F(x, y) = k.]

Short Answer

Step by step solution

(a): Find the orthogonal trajectory of curve 2x2+y2=k

(b): Determine the orthogonal trajectory of curve y=kx4.

(c): Evaluate the orthogonal trajectory of curve y=ekx

(d): Find the orthogonal trajectory of curve y2=kx

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Use the method in Problem 32 to find the orthogonal trajectories for each of the given families of curves, where k is a parameter.(a)2x2+y2=k(b)y=kx4(c)y=ekx(d)y2=kx[Hint: First express the family in the form F(x, y) = k.]

Short Answer

Step by step solution

(a): Find the orthogonal trajectory of curve 2x2+y2=k

(b): Determine the orthogonal trajectory of curve y=kx4.

(c): Evaluate the orthogonal trajectory of curve y=ekx

(d): Find the orthogonal trajectory of curve y2=kx

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Pure Maths

Logic and Functions

Calculus

Decision Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Use the method in Problem 32 to find the orthogonal trajectories for each of the given families of curves, where k is a parameter.
(a) $2 x^{2} + y^{2} = k$
(b) $y = {kx}^{4}$
(c) $y = e^{kx}$
(d) $y^{2} = kx$
[Hint: First express the family in the form F(x, y) = k.]