Question

Another quiz for you:

Find the gradient of a curve.
Write z = -1 + 2i in polar form and exponential form. (or use -1 + 2j if you prefer).
Integrate x sin(x) by parts.
Simplify x^3 + 2 / x^2 + 1.
Solve 5^2x = 125

Answer 1

1. at least the curve should be provided?
2. sorry haven't learnt that yet

Answer 2

I'll give the answers soon.

Answer 3

define gradient, is it the english version or the american version or the surface version?

Answer 4

oh, and for the record; this post makes absolutely no sense ... lol

Answer 5

Stop nitpicking!

Answer 6

3.
\[\int\limits x sinx dx \]\[=-\int\limits x d(cosx)\]\[=-[xcosx - \int\limits cosx d(x)] = sinx - xcosx +C\]

Answer 7

1) Find the gradient of a curve.


2) Write z = -1 + 2i in polar form and exponential form. (or use -1 + 2j if you prefer).


3) Integrate x sin(x) by parts.


4) Simplify x^3 + 2 / x^2 + 1.


5) Solve 5^2x = 125


youve posted 5 different questions under the guise that they all relate somehow

Answer 8

4. x^3 + 2 / x^2 + 1
do you mean x^3 + (2 / x^2) + 1 or (x^3 + 2) / (x^2 + 1) ?

Answer 9

5. 5^2x = 125
i assume it is 5^(2x) = 125
5^(2x) = 125
5^(2x) = 5^3
2x = 3
x = 3/2 = 1.5

Answer 10

For Q4, I mean the latter. Sorry, I should have put it in equation form!

Answer 11

Stop nitpicking, @amistre64!

Answer 12

im not nitpicking, im just trying to understand the question :)

but callisto seems to have parsed it adequately

Answer 13

Nope, he hasn't! He's got Qs 1 and 2 wrong, I'll let him off for saying he hasn't done it yet.

Answer 14

First i'm a she not a he

Answer 15

Second, what i learn for'gradient' is about the slope, if you meant the other thing, of course i can't answer you

Answer 16

Third, i don't think it is possible to do something i haven't learnt

Answer 17

i'm still working on Q4

Answer 18

Sorry Callisto, I apologise sincerely for my actions!

Answer 19

Also, I do mean the slope rather than the other thing. What is the other thing?

Answer 20

how do i know? Slope of a curve = dy/dx , where y is the equation of the curve

Answer 21

(x^3 + 2) /( x^2 + 1) = [x(x^2 +1) -x+2] / (x^2+1) = x - [(x-2)/(x^2 +1)], not sure if this is the answer

Answer 22

actually how are the related?

Answer 23

The answer to question 1 should be that there is no fixed gradient for a curve and you have to differentiate to find the gradient function.

Answer 24

*they

Answer 25

Slope of a curve = dy/dx , where y is the equation of the curve <-- is this correct then?

Answer 26

Yes, sorry for any confusion.

Answer 27

what about the answers to the other questions?

Answer 28

Look at http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/116,117/PolarAndExponentialFormsOfComplexNumbers.pdf
for help with question 2.
Questions 3 and 5 are correct as you posted the correct answers and working.

Answer 29

The working for Question 4 is as follows:
\[(x^3+2)/(x^2+1) = (x(x^2+1) +1-x)/(x^2+1) = x + (1-x)/(x^2+1)\]
Think about what you have to do to the denominator to get the numerator.

Answer 30

no the second step is wrong!
it should be (x(x^2 +1) +2-x)/(x^2+1) !!
check your question again!
i also did the same but i use long division to do so

Answer 31

Yes you're right! :)
Thanks.

Answer 32

thanks, can you teach me how to do the second question?

Answer 33

Complex numbers can be represented in 3 different forms.

These are cartesian, exponential and polar.

The question says to convert -1 + 2i into polar and exponential forms.

This can be represented graphically on an Argand diagram, with the real part on the x axis and the imaginary part on the y axis. So the number will be in the fourth quadrant of the graph.

Complex numbers in polar form have a modulus r and an argument, \[\theta\]. The modulus r is calculated by using Pythagoras' theorem so r = \[\sqrt{2^2 + 1^2} = \sqrt{4 + 1}\] = \sqrt{5}\]

Polar co-ordinates can be converted to Cartesian form by taking x = r cos \[\theta\] and y = r sin \[\theta\] and we need to find the hypoteneuse of the triangle. So we need to use the arctan function (or tan^-1 if you prefer). \[\theta = \arctan (y/x) = \arctan (-2) = -63.43494882\]. However, this is not the right angle. This angle is in the fourth quadrant and the point is in the second quadrant. This is 180 degrees away from the angle so we need to add 180 to the angle we found. The angle between the positive real axis and the point is \[\180 - 63.43... = 116.5650512 = 116.5 = \theta\].

This can be converted to radians accordingly.

To convert this number into exponential form, we need the identity \[z = re^{i \theta}\]. This can be obtained from De Moivre's Theorem. We have r and \[\theta\] so we can convert the number into exponential form. It is best to convert the number into radians before proceeding.

Answer 34

so, for polar, is it just like the polar coordinates but specifically in quad. IV?

Answer 35

What do you mean?

Answer 36

in cartesian form, is it like the ordinary x,y plane but the real no. as x and i as y axis?
in polar form , is it similar to the polar coordinates but the point should lie on quad IV?

Answer 37

The complex number was in cartesian form already, z = x + iy.
Polar form gives it in terms of r and theta rather than x and y. The value of r is unique to the complex number but the argument is not because you could have two complex numbers with the same value of r but different arguments. Take 1 + 2i and 2 + i for example. Both have a modulus of root 5 but the arguments are defined by arctan(2) and arctan(1/2) respectively. Plotting these two complex numbers, or any complex numbers with the same modulus on an Argand diagram should illustrate my case.
The argument of a complex number is the angle between the positive real axis and the line connecting the origin to the point.

Answer 38

got it, thanks!

Answer 39

My pleasure!
If you want me to go through anything else for you, I would be more than happy to do so! (By the way, this is not just for you, this is for everyone I help!)

Answer 40

of course you can, if you want, i bet someone would read and learn something from this post!

Answer 41

That's true.
Please put an image up of you so I can put a face to a name!
Thanks.

Answer 42

i'll consider it seriously as i don't want to be called a 'he' or 'him' again!

Answer 43

Now I know that you are a "she", I will never call you "he" again, I promise!

Answer 44

Callisto does not sound like a "he" or a "she" but I probably should have put that specific post in a better way, I didn't want to say your name too many times, that's what it was!