Twinkle, Twinkle, Little Star (Word by Word Reading) - With Word Meanings

       

Twinkle, twinkle, little star!
How I wonder what you are!
Up above the world so high,
Like a diamond in the sky.  

When the blazing sun is gone,
When he nothing shines upon,
Then you show your little light,
Twinkle, twinkle, all the night.  

Then the traveller in the dark,
Thanks you for your tiny spark,
He could not see which way to go,
If you did not twinkle so.  

In the dark blue sky you keep,
And often through my curtains peep,
For you never shut your eye,
Till the sun is in the sky. 

It's your bright and tiny spark,
Lights the traveller in the dark,
Though I know not what you are,
Twinkle, twinkle, little star. 
    

Drag Me!

Tags: Nursery Rhymes,English Lessons,Communication Skills,

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अध्याय चुनें:

इन वाक्यों को हिंदी से अंग्रेजी में बदलें:

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Some text in the modal.

What is the differentiation and integration of f(x)=1 wrt dx

If $f(x) = 1$, then the derivative of $f(x)$ with respect to $x$ (denoted as $f'(x)$ or $\frac{df}{dx}$) is zero because the function is constant.

$f'(x) = \frac{d}{dx}(1) = 0$

The integral of $f(x) = 1$ with respect to $x$ (denoted as $\int f(x)dx$) is simply $x$ plus a constant of integration (often denoted as $C$).

$\int 1 \,dx = x + C$

So, in summary: $f'(x) = 0$
$\int 1 \,dx = x + C$

where $C$ is the constant of integration.

Change in limits of integration when changing from cartesian to polar coordinates

While changing variables from x, y to r, theta in a double integral, if original limits go from -infinite to +infinite, then what would new limits go to? Please explain step by step.

When changing variables from Cartesian coordinates (x, y) to polar coordinates (r, θ) in a double integral, the limits of integration need to be adjusted accordingly. If the original limits for x and y extend from negative infinity to positive infinity, it typically implies that you are integrating over the entire plane. In polar coordinates, the corresponding region would be described using the radial coordinate (r) and the angular coordinate (θ).

Let's go through the process step by step:

Original double integral:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \,dx\,dy$

Now, let's express the Cartesian coordinates (x, y) in terms of polar coordinates (r, θ):

$x = r \cos(\theta)$ $y = r \sin(\theta)$

Next, we need to express the differential area element $dx\,dy$ in terms of polar coordinates. The area element in Cartesian coordinates is $dx\,dy$, and in polar coordinates, it becomes $r\,dr\,d\theta$. Therefore:

$dx\,dy = r\,dr\,d\theta$

Now, substitute these expressions into the original integral:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \,dx\,dy = \int_{0}^{2\pi} \int_{0}^{\infty} f(r \cos(\theta), r \sin(\theta)) \,r\,dr\,d\theta$

In polar coordinates, the limits for $r$ typically start from 0 to $\infty$, and the limits for $\theta$ range from 0 to $2\pi$ to cover the entire plane.

So, when changing variables from x, y to r, θ in a double integral where the original limits are from $-\infty$ to $+\infty$, the new limits become:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \,dx\,dy$

becomes

$\int_{0}^{2\pi} \int_{0}^{\infty} f(r \cos(\theta), r \sin(\theta)) \,r\,dr\,d\theta$