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Archive for the ‘Calculus’ Category

Confounding Calculus

Another fun Calculus problem we ran into the other day involved using the quotient rule versus using the product rule.  This problem also came from the Briggs and Cochran book in the section on Implicit Differentiation (pg. 162 #15).  So, the problem this time started out as

Use implicit differentiation to find \frac {dy}{dx} .

x^3=\frac {x+y}{x-y}

So, if we start out by differentiating the expressions on each side, we get

3x^2=\frac {(x-y)(1+y')-(x+y)(1-y')}{(x-y)^2}

If we expand and simplify a little –

3x^2=\frac {(x+xy'-y-yy')-(x-xy'+y-yy')}{(x-y)^2}

3x^2=\frac {x+xy'-y-yy'-x+xy'-y+yy'}{(x-y)^2}

3x^2=\frac {2xy'-2y}{(x-y)^2}

So then

3x^2(x-y)^2=2xy'-2y

and

\frac{3x^2(x-y)^2+2y}{2x}=y'

This is the answer that appears in the text – but, if we assume that x\neq y , then we can clear the denominator in the original problem and use implicit differentiation on the result.

x^3=\frac {x+y}{x-y}

x^3(x-y)=x+y

x^4-x^3y=x+y

Then, differentiating

4x^3-(3x^2y+x^3y')=1+y'

or

4x^3-3x^2y-x^3y'=1+y'

Then, to isolate y' :

4x^3-3x^2y-1=x^3y'+y'

4x^3-3x^2y-1=y'(x^3+1)

and

\frac{4x^3-3x^2y-1}{x^3+1}=y'

I spent some time playing around with these, and believe me, they are not the same.  I graphed them and here is what the first answer \frac{3x^2(x-y)^2+2y}{2x}=y' looks like:

And here is the second version, \frac{4x^3-3x^2y-1}{x^3+1}=y'

I tried plugging values for x and y into these different expressions for the derivative of the same function to see if I got the same answer.  This led me to realize the solution to the conundrum!

I started by plugging in some simple values for x and y like x=1, y=1 but quickly realized that this wasn’t valid because x\neq y .  So then I tried x=1, y=0 and for these values, the two expressions were the same!

\frac{3(1)(1-0)^2+2(0)}{2(1)}=y'

\frac{3}{2}=y'

and

\frac{4(1)^3-3(1)^2(0)-1}{(1)^3+1}=y'

\frac{4-1}{1+1}=y'

\frac{3}{2}=y'

These are the same – because the point (1,0) satisfies the original function: x^3=\frac {x+y}{x-y} .

So I got out the Excel spreadsheet and found a series of points that satisfied the original function and tried a few of these points in the two expressions for the derivative and for each one the value of the two expressions is the same!

So next I was determined to show that assuming the original function was true would show the equality of the two different expressions for the derivative.

So, assuming x^3=\frac {x+y}{x-y} we need to turn \frac{4x^3-3x^2y-1}{x^3+1}=y' into \frac{3x^2(x-y)^2+2y}{2x}=y' .

First I substituted \frac{x+y}{x-y} in for x^3 , so:

\frac{4x^3-3x^2y-1}{x^3+1}=y'

\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{x+y}{x-y}+1}=y'

\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{x+y}{x-y}+\frac{x-y}{x-y}}=y'

or

\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{2x}{x-y}}=y'

Then I cleared the fraction in the numerator and denominator:

(\frac{x-y}{x-y})\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{2x}{x-y}}=y'

\frac{4(x+y)-3x^2y(x-y)-1(x-y)}{2x}=y'

OK, so this is

\frac{3x^2y^2-3x^3y+4x+4y-x+y}{2x}=y'

or

\frac{3x^2y^2-3x^3y+3x+5y}{2x}=y'

I saw that where I was headed with this had a 2y and a 3x^2 in it, so I moved a few things around and ended up with:

\frac{3x^2y^2-3x^3y+3(x+y)+2y}{2x}=y'

Going back to the original function x^3=\frac {x+y}{x-y} I saw that I could substitute x^3(x-y) in for the (x+y) :

\frac{3x^2y^2-3x^3y+3(x^3(x-y))+2y}{2x}=y'

\frac{3x^2y^2-3x^3y+3(x^4-x^3y)+2y}{2x}=y'

\frac{3x^2y^2-3x^3y+3x^4-3x^3y+2y}{2x}=y'

Factor out 3x^2 from the everything except the 2y and this is starting to look good:

\frac{3x^2(y^2-xy+x^2-xy)+2y}{2x}=y'

And there it is:

\frac{3x^2(x-y)^2+2y}{2x}=y'

– sigh of relief –

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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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Area Under a Parabola

I was giving a test in my Beginning Algebra classes this past week, and while I was proctoring, I began to page through John Coburn’s Algebra and Trigonometry textbook.  I came across an interesting picture indicating a formula to find the area under a parabola.

parabola area

So, the formula indicates that to find the area under a parabola when it is cut by a horizontal line, we simply multiply two-thirds by the product of the length of the line segment between the points of intersection and the distance from the horizontal line to the vertex.

My first instinct was to see if I could derive this formula myself.

So, I started with a general parabola and, for the sake of simplicity, I put the left intersection point at the origin and made the horizontal line the x-axis.  These choices don’t affect the generality of the derivation.

general parabola

The first thing I did was to integrate the generalized parabola

∫(a(x-h)²+k)dx      from 0 to 2h

This results in an answer of (2/3)ah³+2hk.

If I use the formula (2/3)(2h)(k) I get (4/3)hk.  I was confused and couldn’t see right away why they were different so I integrated the parabola as ax²+bx+c instead to see what would happen.

Using the assumptions, if we integrate

∫(ax²+bx+c)dx      from 0 to 2(-b/2a)=-b/a

We get (b³/6a²) -(bc/a).

If we use the formula using this method, we get

(2/3)(-b/a)(b²/4a-b²/2a+c)=(b³/6a²) -(2bc/3a).

Again, different.

So, I looked at the assumption that the parabola goes through the origin.

If this is true in the first case (where y=a(x-h)²+k), then

0=ah²+k

-k=ah²

Plugging back into the result I had from the first go-round

A=(2/3)ah³+2hk

A=(2/3)(-k)(h)+2hk

A=(4/3)(hk)

Which is what I got using the formula. OK!

Next, the ax²+bx+c method.  For a parabola with the point of intersection at the origin, c=0.

So,

A=(b³/6a²) -(bc/a)

A=(b³/6a²) -(b*0/a)

A=(b³/6a²) -0

A=(b³/6a²)

Using the formula, we got

A=(b³/6a²) -(2bc/3a), but with c=0

A=(b³/6a²) -(2b*0/3a)

A=(b³/6a²) -0

A=(b³/6a²)

So, again they reconcile nicely.  I was glad I realized this quickly otherwise it would have stuck with me for the rest of the day!

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Roots of Polynomials

Rene Descartes determined that if there existed rational roots for a polynomial with integer coefficients, then these roots would be related to the leading coefiicient and constant term in the manner stated in the Rational Roots Theorem.  When I was in high school in the early 80s, we used this theorem to determine all possible rational roots and then used synthetic division to determine which ones actually were roots.  From there, you can decompose the polynomial into factors and try to determine any complex roots.

I still teach the Rational Roots Theorem, but make allowances for today’s technology.  Instead of asking the students to determine all possible rational roots and test them out, we graph the polynomial and use the calculator to determine the rational roots.  Then, we use these roots to break the polynomial down into prime factors, which can allow the students to determine complex roots that did not appear on the graph (I sometimes refer to the complex roots as Sir Not Appearing in This Film).

What if the roots aren’t rational? and other complications

The polynomials I use for exercises, quizzes and tests are set up so that each breaks down to a series of linear factors and one quadratic factor with complex roots, with all the factors having integer coefficients.

But – what if this isn’t the case, or what if there are irrational real roots?  How can we find the complex roots of a polynomial in these situations?

Answer: Newton’s Method.  Newton’s Method is fairly easy to understand in application to finding real roots on the Cartesian Plane, but it also works just as well to find complex roots.  Choosing various real and complex seed values for Newton’s Method leads to finding both the real and the complex roots.

Something very interesting about this process is that if we color the seed values from the complex plane based on which root they eventually lead to, the colored depiction of the complex plane that is produced appears fractal in nature.  Each collection of seed values that leads to the same root is said to lie in a particular Newton Basin.

Here is a great website on Newton Basins.

And another one here.

Here is a look at how they approach this topic at MIT.

(Answer: with technology – Mathematica)

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