   Chapter 4.5, Problem 56E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# In the theory of relativity, the energy of a particle is E = m 0 2 c 4 + h 2 c 2 / λ 2 where m0 is the rest mass of the particle, λ is its wave length, and h is Planck’s constant. Sketch the graph of E as a function of λ. What does the graph say about the energy?

To determine

To Sketch: The graph of energy of particle E as a function of wavelength λ

Explanation

Given:

The given Energy of a particle is E=f(λ)=m02c4+h2c2λ2 (1)

Calculation:

Assume a=m02c4 and b=h2c2

Rewrite the equation (1) with the assumed variable.

E=f(λ)=a+bλ2=aλ2+bλ2=aλ2+bλ

Calculate the asymptotes.

Substitute 0, ± for λ in the above equation by applying limit.

limx0+f(v)=limx0+aλ2+bλ=a(02)+b0=

Therefore, the equation λ=0 is the vertical asymptote.

limxf(v)=limxaλ2+bλ=limxaλ2+bλλλ

limxf(v)=limxa+b/λ21=a+b/21=a+01=a

Therefore, the equation E=a=m0c2 is the horizontal asymptote

Differentiate the equation aλ2+bλ with respect to λ by applying UV method.

f(v)=aλ2+bλf(v)=vu'uv'v2f'(v)=[λ(12(aλ2+b)12)(2aλ)][(aλ2+b)(1)]λ2=aλ2(aλ2+b)12(aλ2+b)12λ2=(aλ2+b)12[aλ2(aλ2+b)]λ2=bλ2aλ2+b<0

Therefore, the function f is decreasing on (0,)

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