   Chapter 10.5, Problem 101E ### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

#### Solutions

Chapter
Section ### Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

# The next two exercises are applications of Einstein’s Special Theory of Relativity and relate to objects that are moving extremely fast. In science fiction terminology a speed of warp 1 is the speed of light-about 3 × 10 8 meters per second. (For instance, a speed of warp 0.8 corresponds to 80% of the speed of light-about 2.4 × 10 8 meters per second.)Lorentz Contraction According to Einstein’s Special Theory of Relativity, a moving object appears to get shorter to a stationary observer as its speed approaches the speed of light. If a spaceship that has a length of 100 meters at rest travels at a speed of warp p, its length in meters, as measured by a stationary observer, is given by L ( p ) = 100 1 − p 2 With domain [ 0 , 1 ] . Estimate L ( 0.95 ) and L ′ ( 0.95 ) . What do these figures tell you?

To determine

To calculate: The value of L(0.95) and L(0.95) and explain it where L(p) represents the length of spaceship from the perspective of observer who is stationary and p is the speed of warp and spaceship length is 100m at rest. The function is L(p)=1001p2 here 0p<1. The speed of warp 1 is the speed of light.

Explanation

Given information:

The expression L(p)=1001p2 represents the length of spaceship in perspective of observer who is stationary and p is the speed of warp and spaceship length is 100m at rest and 0p<1. The speed of warp 1 is the speed of light.

Formula used:

The derivative of a function f(x) using balance difference quotient at any point a is f(a)f(a+h)f(ah)2h

Calculation:

Consider the expression, L(p)=1001p2 represents the length of spaceship from the observer’s perspective.

Evaluate the value of L(0.95) by substituting p=0.95 in expression L(p)=1001p2

L(p)=1001p2L(0.95)=1001(0.95)231.2

Evaluate the value of L(0.95),

L(a)L(a+h)L(ah)2hL(0.95)L(0.95+0.001)L(0

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