# Thread: Having issues with algebra pertaining to a rate of change problem

1. ## Having issues with algebra pertaining to a rate of change problem

Hi and thank you for your help, it is not always easy to teach yourself math.

A motorboat coasts towards a dock with its engine off. It's distance s (in metres) from dock t seconds after the engine is cut is s(t)= {90/[3+t]} - 10, 0<t<6.

I have to find the average velocity during an interval of time {delta}t starting at t=1.

I understand the difference quotient this one being {delta}s over {delta}t.
Such as {delta}t equals 1+{delta}t-1 equals {delta}t

And

{delta}s equals [90/(3+(1+{delta}t) - 10] - [90/3+1]

I know the answer is -[90/(4(4+{delta}t)

My problem is the algebra that gets me to the result of {delta}s for the difference quotient and then to actual difference quotient to get the answer -[90/(4(4+{delta}t)

Thank you, your help is appreciated.

I like math and I am trying to teach myself but sometimes I get stuck.

2. ## Re: Having issues with algebra pertaining to a rate of change problem

Okay, you have $s(1+ \Delta t)- s(1)= \frac{90}{3+ (1+\Delta t)}- 10]- [\frac{90}{3+ 1)}- 10]= \frac{90}{4+\Delta t}- \frac{90}{4}$ as the change in distance.
To subtract those fractions, get a "common denominator" by multiplying numerator and denominator of first fraction by 4 and the numerator and denominator of the second fraction by $4+ \Delta t$:
$\frac{360}{4(4+\Delta t)}- \frac{360- 90\Delta t}{4(4+ \Delta t)}= \frac{90\Delta t}{16+ 4\Delta t}$
The average speed over that time is that change in distance divided by $\Delta t$:
$\frac{90\Delta t}{16\Delta t+ 4(\Delta t)^2}= \frac{90}{16+ 4\Delta t}$

3. ## Re: Having issues with algebra pertaining to a rate of change problem

Ah okay! Thank you. One question though, the books question ended up being negative but yours ended up be positive. Why is that?

4. ## Re: Having issues with algebra pertaining to a rate of change problem

Wrong post, sorry