# Derivative definition problem

## Homework Statement

Suppose that an amount function ## a(t) ## is differentiable and satisfies the property
## a(s + t) = a(s) + a(t) − a(0) ##
for all non-negative real numbers ## s ## and ## t ##.

(a) Using the definition of derivative as a limit of a difference quotient, show that ## a'(t) = a'(0) ##.

(b) Show that ## a(t) = 1 + it ## where ## i = a(1) − a(0) = a(1) − 1 ##.

N/A

## The Attempt at a Solution

I do not understand what part b. expects me to do. If ## a'(t) = a'(0) ##, then I can show that equivalency using the definition of ## i ##. But, does that really show that ## a(t) = 1 + it ##? Perhaps the question is poorly worded, and it should read ## a(t) ## is a possible solution? Or am I looking at this the wrong way?

pasmith
Homework Helper
Any function of the form $a(t) = Ct + D$ for constants $C$ and $D$ satisfies $a(s + t) = a(s) + a(t) - a(0)$ for all nonnegative $s$ and $t$.

Perhaps the definition of an "amount function" imposes conditions on $a$ which you haven't told us about, for example that $a(0) = 1$.

Stephen Tashi
$a'(t) = a'(0)$ implies $a'(t)$ is a constant function. You know how to find an antiderivative of a constant function.

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Suppose that an amount function ## a(t) ## is differentiable and satisfies the property
## a(s + t) = a(s) + a(t) − a(0) ##
for all non-negative real numbers ## s ## and ## t ##.

(a) Using the definition of derivative as a limit of a difference quotient, show that ## a'(t) = a'(0) ##.

(b) Show that ## a(t) = 1 + it ## where ## i = a(1) − a(0) = a(1) − 1 ##.

N/A

## The Attempt at a Solution

I do not understand what part b. expects me to do. If ## a'(t) = a'(0) ##, then I can show that equivalency using the definition of ## i ##. But, does that really show that ## a(t) = 1 + it ##? Perhaps the question is poorly worded, and it should read ## a(t) ## is a possible solution? Or am I looking at this the wrong way?

The conclusion is false. Try ##a(t) = mt## for any nonzero constant ##m##. It satisfies the hypotheses but not the conclusion.

The textbook writes True, True for the solutions, for whatever that's worth.

My approach was:
Since ## a'(t) = a'(0) ##, ## a(t) = a(0) = 1 ##. Then ## a'(t) = a(1) - 1 = 0 = a'(0) ##.

LCKurtz
Homework Helper
Gold Member
The textbook writes True, True for the solutions, for whatever that's worth.

My approach was:
Since ## a'(t) = a'(0) ##, ## a(t) = a(0) = 1 ##. Then ## a'(t) = a(1) - 1 = 0 = a'(0) ##.
No. Since ## a'(t) = a'(0) ## then ##a(t) = ta'(0) + C##, and you aren't given ##a(0)=1##.

No. Since ## a'(t) = a'(0) ## then ##a(t) = ta'(0) + C##, and you aren't given ##a(0)=1##.

My bad. ## a(0) = 1 ## for accumulation functions.

LCKurtz
Homework Helper
Gold Member
No. Since ## a'(t) = a'(0) ## then ##a(t) = ta'(0) + C##, and you aren't given ##a(0)=1##.
My bad. ## a(0) = 1 ## for accumulation functions.

Accumulation functions? Who said anything about accumulation functions, whatever they are? Not good to keep secrets when stating a problem...

Accumulation functions? Who said anything about accumulation functions, whatever they are? Not good to keep secrets when stating a problem...

Miswrote, meant amount function as specified in problem. And sorry, I was lazy and assumed too much of whoever was going to help me.

vela
Staff Emeritus