Homework: Function and Graded Problems Math
Solutions to Graded Problems Math 200 Section 1.6
Homework 2
September 17, 2010
20. In the theory of relativity, the mass of a particle with speed v is m = f (v) = m0 1 − v 2 /c2
where m0 is the rest mass of the particle and c is the speed of light in a vacuum. Find the inverse function of f and explain its meaning. Solution. We simply solve for v: m= m0 1− v 2 /c2 =⇒ m 1 − v 2 /c2 = m0
=⇒ m2 1 −
v2 c2
= m2 0
m2 v2 =⇒ 1 − 2 = 0 c m2 =⇒ v2 m2 =1− 0 c2 m2 m0 m m0 m
2
=⇒ v 2 = c2 1 −
2
=⇒ v = ±c 1 −
Our new function v(m) gives velocity v as a function of m. In particular, v(m) gives the velocity (as measured by a relatively stationary observer) that an object of rest mass m0 must be traveling at to have mass m. Section 2.2 14. Sketch the graph of an example of a function f that satisfies all of the given conditions: x→0− lim f (x) = 1, lim f (x) = 1,
x→0+
lim f (x) = −1,
x→2−
lim f (x) = 0,
x→2+
f (2) = 1,
f (0) is undefined
38. (a) Evaluate h(x) = (tan x − x)/x3 for x = 1, 0.5, 0.1, 0.05, 0.01, and 0.005. Solution. Using a calculator, one obtains h(1) = 0.5574 1
Solutions to Graded Problems Math 200
Homework 2
September 17, 2010
h(0.5) = 0.3704 h(0.1) = 0.3347 h(0.05) = 0.3337 h(0.01) = 0.3334 h(0.005) = 0.3333 Solution. Here is an example of a graph that works. Note that there are many solutions. y
x
(b) Guess the value of limx→0 Solution.
tan x−x . x3 tan x−x x3
Based on part (a), we guess that limx→0
= 1. 3
(c) Evaluate h(x) for successively smaller values of x until you finally reach a value of 0 for h(x). Are you still confident that your guess in part (b) is correct? Explain why you eventually obtained a value of 0. (In Section 4.4 a method for evaluating the limit will be explained.) Solution. Note that h(0.00001) = 0. This is surprising, and we no longer have faith in our answer for part (b). Note that we can break up h in the following way tan(x) − x tan(x) x = + 3 3
Homework 2
September 17, 2010
20. In the theory of relativity, the mass of a particle with speed v is m = f (v) = m0 1 − v 2 /c2
where m0 is the rest mass of the particle and c is the speed of light in a vacuum. Find the inverse function of f and explain its meaning. Solution. We simply solve for v: m= m0 1− v 2 /c2 =⇒ m 1 − v 2 /c2 = m0
=⇒ m2 1 −
v2 c2
= m2 0
m2 v2 =⇒ 1 − 2 = 0 c m2 =⇒ v2 m2 =1− 0 c2 m2 m0 m m0 m
2
=⇒ v 2 = c2 1 −
2
=⇒ v = ±c 1 −
Our new function v(m) gives velocity v as a function of m. In particular, v(m) gives the velocity (as measured by a relatively stationary observer) that an object of rest mass m0 must be traveling at to have mass m. Section 2.2 14. Sketch the graph of an example of a function f that satisfies all of the given conditions: x→0− lim f (x) = 1, lim f (x) = 1,
x→0+
lim f (x) = −1,
x→2−
lim f (x) = 0,
x→2+
f (2) = 1,
f (0) is undefined
38. (a) Evaluate h(x) = (tan x − x)/x3 for x = 1, 0.5, 0.1, 0.05, 0.01, and 0.005. Solution. Using a calculator, one obtains h(1) = 0.5574 1
Solutions to Graded Problems Math 200
Homework 2
September 17, 2010
h(0.5) = 0.3704 h(0.1) = 0.3347 h(0.05) = 0.3337 h(0.01) = 0.3334 h(0.005) = 0.3333 Solution. Here is an example of a graph that works. Note that there are many solutions. y
x
(b) Guess the value of limx→0 Solution.
tan x−x . x3 tan x−x x3
Based on part (a), we guess that limx→0
= 1. 3
(c) Evaluate h(x) for successively smaller values of x until you finally reach a value of 0 for h(x). Are you still confident that your guess in part (b) is correct? Explain why you eventually obtained a value of 0. (In Section 4.4 a method for evaluating the limit will be explained.) Solution. Note that h(0.00001) = 0. This is surprising, and we no longer have faith in our answer for part (b). Note that we can break up h in the following way tan(x) − x tan(x) x = + 3 3