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A geometric proof of trigonometric derivatives

May 19, 2009 5 comments

As we already know, the first derivatives of \sin(x) and \cos(x) are \cos(x) and -\sin(x) respectively. The question is how do we prove it?

Since high school, the only proof that I knew is to use two of the trigonometric identities which are:

  • \sin \alpha - \sin \beta = 2 \sin (\frac{\alpha-\beta}{2}) \cos (\frac{\alpha+\beta}{2}), and
  • \cos \alpha - \cos \beta = -2 \sin (\frac{\alpha-\beta}{2}) \sin (\frac{\alpha+\beta}{2}).

Hence,

\frac{d}{dx}\sin x = \displaystyle{\lim_{\Delta x \rightarrow 0} \frac{\sin(x+\Delta x)-\sin(x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} 2 \sin(\frac{\Delta x}{2})\cos(\frac{2x+\Delta x}{2})=\cos x}.

The derivative of \cos x can also be derived using the second identity.

However, there is another proof, which is (I think) more elegant. It is using a geometric interpretation of \sin x and \cos x. Let us draw a unit circle in \mathbb{R}^2.

geo_proof1

Thus we have y=\sin \theta and x=\cos \theta. We are looking for \frac{dy}{d \theta} and\frac{dx}{d \theta}, however we are not going to discuss the first derivative of \cos \theta as it can be done in the same way as \sin \theta.

\displaystyle{\frac{d \sin \theta}{d \theta}=\lim_{\Delta \theta \rightarrow 0} \frac{\sin(\theta + \Delta \theta)-\sin(\theta)}{\Delta \theta}=\lim_{\Delta \theta \rightarrow 0} \frac{y+\Delta y - y}{\Delta \theta}=\lim_{\Delta \theta \rightarrow 0} \frac{\Delta y}{\Delta \theta}}.

We are now going to estimate \Delta y, see the second figure.  geo_proof2The following relationship applies:

\sin \phi = \frac{\Delta y}{r} . or \Delta y = r \sin \phi.

Thus,

\displaystyle{\frac{d \sin \theta}{d \theta}=\lim_{\Delta \theta \rightarrow 0} \frac{\Delta y}{\Delta \theta}=\lim_{\Delta \theta \rightarrow 0} \sin \phi \frac{r}{\Delta \theta}},

where r is the chord PQ and \Delta \theta is the arc PQ. As \Delta \theta goes to zero, the ratio of the arc PQ and the chord PQ tends to 1 and the chord PQ tends to become a tangent line of the unit circle at (x,y), therefore \sin \phi = \sin (\theta + \pi/2). Assuming that the limit of multiplication is the same as the multiplication of limit, we have the following:

\frac{d \sin \theta}{d \theta}=\sin(\theta + \pi/2) = \cos(\theta).

As mentioned before, the first derivative of \cos \theta can be done the same way.

Also, I should thank Pak Yudi Soeharyadi as I got the idea of this geometric proof from him.

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Fundamental Theorem of Calculus is fundamental !

May 11, 2009 1 comment

Most of my students do not seem to understand the two concepts of integrals. I have been there as well, so I did not blame them for not understanding the concept of integrals.  Even I sometimes forget. From that reason I try to post what the fundamental theorem of calculus means exactly.

It is all started with the definition first definition of integrals, which is formulated by the famous mathematician Bernhard Riemann. This is called the definite integral, or \int_a^b f(x) dx .

We can think that \int_a^b f(x) dx as the area below the curve f(x), see the picture below.

kalkulusThus,\int_a^b f(x) dx is the area S. As a first approximation, we shall divide the interval (a,b) into an equal number of segments, say \Delta x. Then we can approximate the area of S as follows:

\int_a^b f(x) dx \approx \sum_{i=0}^n f(x_i) \Delta x.

The right hand side of the equation above is called the Riemann sum. Our final step to find the value of definite integral is to take limit the Riemann sum as \Delta x \rightarrow 0,

\int_a^b f(x) dx = \lim_{\Delta x \rightarrow 0} \sum_{i=0}^n f(x_i) \Delta x.

Now, the second concept about integrals will be explained. This concept is called the indefinite integral or is also usually called the anti-derivative. Suppose F,f are functions of x such that F'(x)=f(x) (or we can say that the derivative of F(x) is f(x) or, the anti-derivative of f(x) is F(x)), then

\int f(x) dx = F(x) + C,

where C is a constant real number.

Hence, these two are two different concepts that are not related one to another. However, there is a great theorem that relates those two. The theorem perfectly and comfortably links the limit of Riemann sum and the anti-derivative. Actually there are two great theorems.

First Fundamental Theorem of Calculus Suppose f is a continuous function defined on closed interval [a,b] and define F as

F(x) = \int_a^x f(t) dt,

then F is continuous and differentiable, moreover, F'(x) = f(x).

Second Fundamental Theorem of Calculus Supposef is a function defined on closed interval [a,b] and F is the anti-derivative of f. If f is integrable on [a,b], then

\int_a^b f(x) dx = F(b) - F(a).

Therefore, if we want to compute, for examples, \int_0^1 x^2 dx, we don’t have to find the limit of Riemann sum of \sum_{i=0}^n x_i^2 \Delta x as \Delta x goes to zero.

Categories: Justincase, Teachings Tags: ,

Result of mid semester exam 2 of Calculus II (UTS 2)

May 7, 2009 2 comments

As I promised yesterday, you can now see the result of mid semester exam 2 of calculus II.

Please go to this page.

Next wednesday is gonna be our last tutorial. I hope you can do well in the final examination.

Categories: Teachings Tags: , ,