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Chapter 8: Q18P (page 407)

Find the family of orthogonal trajectories of the circles (x-h)2+y2=h2. (See the instructions above Problem 2.31.)

Short Answer

Expert verified

Answer

The family of orthogonal trajectories of the circles:

Step by step solution

01

Given information

The given equation of the circle isx-h2+y2=h2.

02

Standard equation of a circle

The standard equation of a circle is (x-h)2+(y-k)2=r2

Here, (h,k)is the center of the circle and r is the radius of the circle.

03

The orthogonal trajectories of the circle

The given equation of the circle is,

x-h2+y2=h2

The standard equation of a circle is

x-h2+y-k2=r2............i

where h,k is the center of the circle and r is the radius of the circle.

Now, compare equation (1) with the standard form of equation, which indicates that

h,0is the center of the circle and h is the radius of the circle.

Therefore, it can be said that the equation (1) shows a family of circles with center h,0and radius h. The value of h varies from ∞to ∞.

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Most popular questions from this chapter

Find the solutions of (1.2)(putI=dqIdt)and (1.3), if V=V0sinω't( const.).

(a) Show that D(eaxy)=eax(D+a)y,D2(eaxy)=eax(D+a)2y, and so on; that is, for any positive integral n, Dn(eaxy)=eax(D+a)ny.

Thus, show that ifis any polynomial in the operator D, then L(D)(eaxy)=eaxL(D+a)y.

This is called the exponential shift.

(b) Use to show that (D-1)3(exy)=exD3y,(D2+D-6)(e-3xy)=e-3x(D2-5D)y..

(c) Replace Dby D-a, to obtain eaxP(D)y=P(D-a)eaxy

This is called the inverse exponential shift.

(d) Using (c), we can change a differential equation whose right-hand side is an exponential times a

polynomial, to one whose right-hand side is just a polynomial. For example, consider

(D2-D-6)y=10×e2x; multiplying both sides by e-3xand using (c), we get

e-3x(D2-D-6)y=[D+32-D+3-6"]"ye-3x=(D2+5D)ye-3x=10x

Show that a solution of (D2+5D)u=10xis u=x2-25x; then or use this method to solve Problems 23 to 26.

Find the distance which an object moves in time tif it starts from rest and has accelerationd2xdt2=ge−kt. Show that for smalltthe result is approximately, and for very large, the speeddxdtis approximately (1.10)constant. The constant is called the terminal speed . (This problem corresponds roughly to the motion of a parachutist.)

Solve the differential equation yy2+2xy'-y=0by changing from variables role="math" localid="1655272385100" y,xto r,x where y2=r2; then yy'=n'-x.

(a)Consider a light beam travelling downward into the ocean. As the beam progresses, it is partially absorbed and its intensity decreases. The rate at which the intensity is decreasing with depth at any point is proportional to the intensity at that depth. The proportionality constant μis called the linear absorption coefficient. Show that if the intensity at the surface is I0, the intensity at a distance s below the surface is I=I0eμs. The linear absorption coefficient for water is of the order of 10.2ft.1(the exact value depending on the wavelength of the light and the impurities in the water). For this value of μ, find the intensity as a fraction of the surface intensity at a depth of 1 ft, ft,ft,mile. When the intensity of a light beam has been reduced to half its surface intensity (I=12I0), the distance the light has penetrated into the absorbing substance is called the half-value thickness of the substance. Find the half-value thickness in terms of μ. Find the half-value thickness for water for the value of μgiven above.

(b) Note that the differential equation and its solution in this problem are mathematically the same as those in Example 1, although the physical problem and the terminology are different. In discussing radioactive decay, we call λthe decay constant, and we define the half-life T of a radioactive substance as the time when N=12N0(compare half-value thickness). Find the relation between λand T.
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