A analytical approach to the differential equation

[Hummingbird25]

An analytical approach to the differential equation(urgent)

Homework Statement

(1) Given the function [tex]f: t \mapsto |t|[/tex], show that the function f is a differentiable function where [tex]t \neq 0[/tex] and write f's the differential coffiecient.

2) Given the function [tex]g: t \mapsto \sqrt{t^2}[/tex], show that g is differentiable if [tex]t \neq 0[/tex], and write the differentiable coeffient of g.

Homework Equations

Also my textbook asks why can't you use the chainrule to find the differential coefficient at t = 0. Isn't that because the limit doesn't exist?

The Attempt at a Solution

My Solution part(1):

According to the definition of my textbook for a function to be a differentiable function the following must be meet:

(1) Let [tex]A \subseteq \mathbb{R}[/tex], let [tex]f: A \rightarrow \mathbb{R}[/tex] be a real function defined on A, and let 'a' be a point in A.

Then f is a differentiable function at a, if t is an inner point of A and the differential coeffient is

[tex]\frac{f(t)-f(a)}{t-a}[/tex] has a finit limit where t tends towards a.

Therefore [tex]f'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a}[/tex]



Assuming that t = 0, then it an inner point which implies that exist an r > 0 [tex]\Rightarrow B_{r}(t = 0) \in A[/tex]. (Refer to the definition of the inner point). But according of the limit, this doesn't exist then t = 0.

So therefore

[tex]f'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a} = \mathop {\lim }\limits_{t \to a}\frac{|t|-|a|}{t-a}[/tex] is the differential coeffient is required which is only valied at [tex]t \neq 0.[/tex]

Assuming that t = 0, then it an inner point which implies that exist an r > 0 [tex]\Rightarrow B_{r}(t = 0) \in A[/tex]. (Refer to the definition of the inner point). But according of the limit, this doesn't exist then t = 0.

My solution(2)

2) Given the function [tex]g: t \mapsto \sqrt{t^2}[/tex], show that g is differentiable if [tex]t \neq 0[/tex], and write the differentiable coeffient of g.

As in (1) x = 0 is assumed not be an inner point, and therefore its not possible to find the differentiable coeffient at that point?

Then the differentiable coeffient is


[tex]g'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a} = \mathop {\lim }\limits_{t \to a}\frac{|t|-|a|}{t-a}[/tex]

Since [tex]\sqrt{t^2} = |t|[/tex] according to My TI-92 calculator.

Isn't this verified by

[tex]g'(t) = \frac{2t}{2 \cdot \sqrt{t^2}}[/tex] where [tex]g(t) = \int (g'(t)) dt = |t| [/tex]

Cheers Hummingbird

 

[Gib Z]

1) Just use the piecewise definition of the absolute value function;

[tex]|x| = x \mbox{ if } x\geq 0, -x \mbox{ if } x< 0[/tex]. We don't really even need to take any limits when we split it into these two cases. Just show that the derivative is 1 when x> 0, the derivative is -1 when x< 0 and that since the left hand and right hand limits are different, the limit does not exist, ie the derivative does not exist.

2) The easiest way to do this may to be show that [tex]\sqrt{t^2} = |x|[/tex], or otherwise, as you did, use the chain rule to show the derivative is given by [tex]\frac{t}{\sqrt{t^2}}[/tex]. At t=0, that expression has no meaning, ie the derivative does not exist at t=0.

PS - I guess your textbook is ancient, judging from the fact your book uses, aside from a few other signs, the term "differential coefficient", which Courant says is only used in "older" textbooks, and Courant isn't exactly bright and shine new either ! But it doesn't matter, just be happy your learning a rigorous treatment of calculus the first time around. You have no idea how luck you are =]

 

[HallsofIvy]

Homework Statement

(1) Given the function [tex]f: t \mapsto |t|[/tex], show that the function f is a differentiable function where [tex]t \neq 0[/tex] and write f's the differential coffiecient.

2) Given the function [tex]g: t \mapsto \sqrt{t^2}[/tex], show that g is differentiable if [tex]t \neq 0[/tex], and write the differentiable coeffient of g.

Homework Equations

Also my textbook asks why can't you use the chainrule to find the differential coefficient at t = 0. Isn't that because the limit doesn't exist?

The Attempt at a Solution

My Solution part(1):

According to the definition of my textbook for a function to be a differentiable function the following must be meet:

(1) Let [tex]A \subseteq \mathbb{R}[/tex], let [tex]f: A \rightarrow \mathbb{R}[/tex] be a real function defined on A, and let 'a' be a point in A.

Then f is a differentiable function at a, if t is an inner point of A and the differential coeffient is

[tex]\frac{f(t)-f(a)}{t-a}[/tex] has a finit limit where t tends towards a.

Therefore [tex]f'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a}[/tex]



Assuming that t = 0, then it an inner point which implies that exist an r > 0 [tex]\Rightarrow B_{r}(t = 0) \in A[/tex]. (Refer to the definition of the inner point). But according of the limit, this doesn't exist then t = 0.

So therefore

[tex]f'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a} = \mathop {\lim }\limits_{t \to a}\frac{|t|-|a|}{t-a}[/tex] is the differential coeffient is required which is only valied at [tex]t \neq 0.[/tex]

Assuming that t = 0, then it an inner point which implies that exist an r > 0 [tex]\Rightarrow B_{r}(t = 0) \in A[/tex]. (Refer to the definition of the inner point). But according of the limit, this doesn't exist then t = 0.

Have you lost track of what the question asks? You are asked to show that the derivative does exist as long as t is not 0. You have written all of this to show that the derivative does not exist at t= 0 while the problem asks nothing about what happens at t= 0!

My solution(2)

2) Given the function [tex]g: t \mapsto \sqrt{t^2}[/tex], show that g is differentiable if [tex]t \neq 0[/tex], and write the differentiable coeffient of g.

As in (1) x = 0 is assumed not be an inner point, and therefore its not possible to find the differentiable coeffient at that point?

No, it is not "therefore". Again, this question has nothing to do with what happens at t= 0.

Then the differentiable coeffient is


[tex]g'(t) = \mathop {\lim }\limits_{t \to a} \frac{f(t)-f(a)}{t-a} = \mathop {\lim }\limits_{t \to a}\frac{|t|-|a|}{t-a}[/tex]

Since [tex]\sqrt{t^2} = |t|[/tex] according to My TI-92 calculator.

Isn't this verified by

[tex]g'(t) = \frac{2t}{2 \cdot \sqrt{t^2}}[/tex] where [tex]g(t) = \int (g'(t)) dt = |t| [/tex]

Cheers Hummingbird

Yes, since [itex]\sqrt{t^2}= |t|[/itex] (and you shouldn't need a calculator to tell you that!) you can use (1) above. However, I suspect this problem was to show that you get the same thing. You can multiply bothe numerator and denominator of
[tex]\frac{\sqrt{t}- \sqrt{a}}{t- a}[/tex]
by [itex]\sqrt{t}+ \sqrt{a}[/itex] and see what you get.

By the way, this problem is about the derivative. It has nothing to do with differential equations.

 

[qspeechc]

I'd like to know what textbook you use Hummingbird- I think I like the sound of it :P

 

[Hummingbird25]

I'd like to know what textbook you use Hummingbird- I think I like the sound of it :P

Hello Hallsoft and qspeechc,

1)

The function [tex]f: t \mapsto |t|[/tex], show that the function is differentiable where

[tex]t \neq 0[/tex] and write the derivative.

First I look at the cases where t > 0 and t < 0.

The first t > 0, I get the derivative to be

[tex]f'(t) = \frac{|t|-|a|}{t-a} = \frac{t-a}{t-a} = 1[/tex] for all t in [tex]\mathbb{R}^{+}[/tex]

Then the case if t < 0,

[tex]f'(t) = \frac{|t|-|a|}{t-a} = \frac{-t-(-a)}{t-a} = -1[/tex] for all t in [tex]\mathbb{R}^{-}[/tex]

finally show why the derivative of the function cannot be found at t = 0.

at the case where t = 0, I get

[tex]f'(t) = lim_{t \rightarrow 0} \frac{|a|}{a}[/tex] which doesn't exist. Thusly the derivative doesn't exist at t = 0.

I hope I have understood the question correctly now?

Sincerely Yours

Hummingbird.

 

[Hummingbird25]

Hello Hallsoft,

Just to be clear. Its your opinion that I can use the argument in qoute 2. In Both question one and question 2??

Sincerely Yours

Hummingbird.

p.s. if I multiply [tex]\frac{\sqrt{t^2}-\sqrt{a^2}}{t-a} \cdot \sqrt{t}+\sqrt{a} = \frac{\sqrt{t^2}-\sqrt{a^2} \cdot \sqrt{t}+\sqrt{a}}{t-a}[/tex] I am sorry but what does that change?

 

[HallsofIvy]

Sorry, I just noticed that I had miswritten the numerator of your fraction as
[tex]\sqrt{t}- \sqrt{a}}[/tex]
when it should have been
[tex]\sqrt{t^2}- \sqrt{a^2}[/tex]
What I meant was to multiply both numerator and denominator by [itex]\sqrt{t^2}+ \sqrt{a^2}[/itex]

 

[Hummingbird25]

Hello Hall,

I been trying something here

Original question (1) Show that the derivative of function [tex]t \mapsto |t|[/tex] exists at [tex]t \neq 0[/tex] and write the derivative. Next write out the derivative. Finally show why the derivative doesn't exist at t = 0.

to show that the derivative of [tex]t \mapsto |t|[/tex] does not exist at t = 0.

I say that [tex]\frac{f(t)-f(0)}{t-0} = \frac{|t|}{t}[/tex] which is undefined if t = 0.

Thusly the derivative doens't exist at t = 0.

Next I am trying to show that the derivative does exist if [tex] t \neq 0[/tex]

This inturn allows me to write that

[tex]f'(t) = \frac{|t|-|t_{0}|}{t-{t_0}}[/tex] where if [tex]\mathop {\lim }\limits_{t \to t_{0} = 1 } \frac{|1|-|t_0|}{1-t_{t_0}} = 1 [/tex] if t = 1 and if t = - 1 [tex]f'(t) = \frac{|t|-|t_{0}|}{t-{t_0}}[/tex] where if [tex]\mathop {\lim }\limits_{t \to t_{0} = -1 } \frac{|-1|-|t_0|}{-1-t_{0}} = -1 [/tex]. Thusly the derivative exists if [tex]t \neq 0[/tex]. How does it look now?

Sincerely Yours
Hummingbird.