Tag Archives: Calculus for IITJEE

Derivatives: IITJEE Mains Maths: Mathematics Hothouse video lecture

Limits Part 2: IITJEE Mains maths: Mathematics Hothouse

Limits part 1: video lecture: IITJEE Mains maths: Mathematics Hothouse

 

What is a catenary or a chain curve? A nice little application of basic calculus

Let us continue our exploration of basic calculus and its application. As you will discover, the invention of calculus is a triumph of the human intellect, it is a fountain head of many ideas in pure mathematics as well as mine of applications to sciences and engineering and even economics and humanities!

Hanging cables

Imagine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable’s weight per unit length is w and horizontal tension at its lowest part is a vector of length H. If we choose a coordinate system for the plane of the cable in which the x-axis is horizontal, the force of gravity is straight down, the positive y-axis points straight up, and the lowest point of the cable lies at the point y=H/w on the y-axis. (Fig 1), then it can be shown that the cable lies along the graph of the hyperbolic cosine

y=(H/w)\cosh (wx/H).

Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena meaning ‘chain’.

a) Let P(x,y) denote an arbitrary point on the cable. Fig 2 displays the tension at P as a vector of  length (magnitude) T, as well as the tension at H at the lowest point A. Show that the cable’s slope at P is

\tan \phi =dy/dx=\sinh (wx/H)

b) Using the result from part (a) and the fact that the tension at P must equal H (the cable is not moving or swinging), show that T=wy. This means that the magnitude of the tension at P(x,y) is exactly equal to the weight of y units of cable.catenarycurveblog

2) (Continuation of above problem). The length of arc AP in Fig 2 is

s=(1/a)\sinh (ax) where a=w/H. Show that the coordinates of P may  be expressed in terms of s as

x=(1/a) (\sinh)^{-1} (as) and y=\sqrt (s^{2}+1/a^{2}).

3) The sag and horizontal tension in a cable. The ends of a cable 32 feet long and weighing 2 pounds per foot are fastened at the same level in posts 30 feet apart.

i) Model the cable with the equation

y=(1/a)\cosh (ax) for -15 \leq x \leq 15

Use information from problem 2  above to show that a satisfies the equation

16a=\sinh 15a

ii) Estimate the horizontal tension in the cable at the cable’s lowest point.

I hope you enjoy a lot 🙂

More later…

Nalin Pithwa

Implicit differentiation example — Helga von Koch’s snowflake curve (1904)

HelgavonKoch snowflakel

Let us continue further our exploration of IITJEE Calculus. Especially, implicit differentiation.

Start with an equilateral triangle, calling it curve 1. On the middle third of each side, build an equilateral triangle pointing outward. Then, erase the interiors of the old middle thirds. Call the expanded curve curve 2. Now, put equilateral triangles, again pointing outward, on the middle thirds of  the sides of curve 2. Erase the interiors of the old middle thirds to  make curve 3. Repeat the process, as shown, to define an infinite sequence of plane curves. The limit curve of the sequence is Koch’s snowflake curve.

The snowflake curve is too rough to  have a tangent at any point. In other words, the equation F(x,y)=0 defining the curve does not define y as a differentiable function of x or x as a differentiable function of y at any point. We will encounter snowflake again when we study length.

Please download the attached figure.

More later…

Nalin Pithwa

How rough can the graph of a continuous function be?

We know that if a function f(x) has a derivative at x=c, then f(x) is continuous at x=c. But, the converse of this theorem is false. A function need not have a derivative at a point, where it is continuous, as in the following example:

The function y=|x| is differentiable on (-\infty, 0) and

(0,\infty) but has no derivative at x=0.

Also, using this simple idea, we can use a sawtooth graph to define a continuous function that fails to have a derivative at infinitely many points.

But, can a continuous function fail to have a derivative at every point?

The answer, surprisingly enough, is yes, as Karl Weierstrass (1815-1897) found in 1872. One of his formulae (there are many others like it) was

f(x)=\sum_{n=0}^{\infty} (2/3)^{n} \cos (9^{n}\pi x)

a formula that expresses the function f as an infinite sum of cosines with increasingly higher frequencies. By adding wiggles to wiggles infinitely many times, so to speak, the formula produces a graph that is too bumpy in the limit to have a tangent anywhere.

Continuous curves that fail to have a tangent anywhere play a useful role in chaos theory, in part because there is no way to assign a finite length to such a curve. We will see what length has to do with derivatives later in one of the blog articles.

More later…

Nalin Pithwa

a non-trivial limit problem — calculus IIT JEE

Here is a what I call, a non-trivial problem of  evaluation of a limit. (Hey, I call it non-trivial because I could  only find a longish, brute-force solution as given below. If you have a clever, elegant solution, I would like to learn from you. Just write it down, take a snapshot in your smartphone camera, and post it in comments :-)).

Evaluate \lim_{x \rightarrow 0} {((a^{x}+b^{x}+c^{3})/3)}^{2/x}

Solution. The above limit can be recast as follows:

\lim_{x \rightarrow 0}{((a^{x}-1+b^{x}-1+c^{x}-1+3)/3)}^{2/x}

=\lim_{x \rightarrow 0}{((a^{x}-1)/3+(b^{x}-1)/3+(c^{x}-1)/3+1)}^{2/x}

Now, observe as x \rightarrow 0, (a^{x}-1)/3+(b^{x}-1)/3+(c^{x}-1)/3 also tends to zero.

So, let X=((a^{x}-1)/3+(b^{x}-1)/3+(c^{x}-1)/3). Hence, the required limit is equal to

\lim_{X \rightarrow 0}{(1+X)}^{(2/X)(X/x)}

=\lim_{X \rightarrow 0}{{((1+X)}^{(1/X)})}^{2X/x} call this A but we know that

lim_{X \rightarrow 0} {(1+X)}^{1/X}=e

Hence, the above expression becomes e^{(2/3)(( \lim_{x \rightarrow 0}(a^{x}-1)/x)+(\lim_{x \rightarrow 0}(b^{x}-1)/x)+(\lim_{x \rightarrow 0}(c^{x}-1)/x))}

=e^{(2/3)((\ln a)+(\ln b)+(\ln c))}

=e^{(2/3)(\ln (abc))}

={(abc)}^{2/3}.

So, what do you think — is there an alternate elegant solution to  my brute force solution above? if you have, please share it with me also.

More later…

Nalin Pithwa