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Archive for October, 2012

I’m teaching a Calculus course this year and came across an interesting problem the other day.  We were working from the Briggs & Cochran Calculus book and discussing the quotient rule.  Problem #40 on page 127 says given the function:

f(t)=\frac {3t^2}{t^2+1}

a) Find the values of t for which the slope of the curve is zero.

b) Does the graph of f(t) have a slope of 3 at any point?

OK – so we need to apply the quotient rule and find f'(t)

f'(t)= \frac{(t^2+1)(6t)-(3t^2)(2t)}{(t^2+1)^2}

OR

f'(t)= \frac{6t^3+6t-6t^3}{(t^2+1)^2}

f'(t)= \frac{6t}{(t^2+1)^2}

OK – so part (a) is easy enough.  The derivative is equal to zero when x=0 .  But what about part (b) ?  We can graph the derivative and see that it’s always less than 3 , but what about algebraically?

The problem can be posed in a number of different ways.  I chose to set it up as:

3> \frac{6t}{(t^2+1)^2}

so

3(t^2+1)^2 > 6t

and

(t^2+1)^2 > 2t

Again, we can graph this and SEE clearly that the graph of (t^2+1)^2 is definitely above the graph of 2t , but I like to know what’s happening algebraically.  In fact, I used to give my College Algebra students algebraic inequalities as an introductory project, so I really wanted to figure this out.

Of course, this came up in the middle of class, so we looked at the graph and I began to consider the algebraic reasoning, but didn’t get far, so I told the class that I would think about it while they had their quiz – but still to no avail.  I worked on it during my free time yesterday and made some progress, then when I came in this morning it all fell together:

t=0

\frac{6t}{(t^2+1)^2}=0

OK, done there.

t<0

6t<0

(t^2+1)^2>0

\frac{6t}{(t^2+1)^2}<0

Again – ok, done.  What was the sticking point for me yesterday were the positive values of t . So:

t>0

(t-1)^2\geq0

t^2-2t+1\geq0

So, t^2+1\geq2t

Also, t^2+1>1

So, (t^2+1)(t^2+1)>1(t^2+1)\geq2t

So now we know that (t^2+1)^2>2t  and can work our way back to

3> \frac{6t}{(t^2+1)^2}

For some reason, I find this beautiful!

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