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First Year Mathematics A (Differentiation & Linear Algebra) (3 Viewers)

leehuan

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Re: MATH1131 help thread

yeah i want to know if i'm supervising you :(


pls pm
Woooow
Use quadratic formula to find all complex roots:

-2z^2+6z-3


I get the correct answer - except for the i. I don't get an i in my answer. Where does the i come in in this question??

Thanks.
http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=-2z^2+6z-3=0
Yeah just to back what InteGrand said, if you plug your equation in you find that the complex roots are both real
 

Flop21

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Re: MATH1131 help thread

yeah i want to know if i'm supervising you :(


pls pm
I know you want to, but I don't want to say haha.


I think I know who you are tho

but you won't know me


:)
 

Flop21

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Re: MATH1131 help thread

Regarding polar form...

My book says the polar form is z = re^(i*theta). But then the answers use the fact that e^(i*theta) = costheta +isintheta, and put it in that form. How do I know what form they want? Do I always do it in the latter form, where I get rid of e, or the former??
 

leehuan

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Re: MATH1131 help thread

Regarding polar form...

My book says the polar form is z = re^(i*theta). But then the answers use the fact that e^(i*theta) = costheta +isintheta, and put it in that form. How do I know what form they want? Do I always do it in the latter form, where I get rid of e, or the former??
Pretty sure that does not matter. Otherwise they will specify what to put in on the exam.

After all, reix = r(cos(x)+isin(x)) for all real x, r
 

Flop21

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Re: MATH1131 help thread

Why am I wrong here? By looking at the graph there are only 4 points of intersection I can see. The highlighted part of the graph does not intersect, you can see that when you zoom in.

 

InteGrand

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Re: MATH1131 help thread

Why am I wrong here? By looking at the graph there are only 4 points of intersection I can see. The highlighted part of the graph does not intersect, you can see that when you zoom in.

Have you typed the equations correctly? For the first one, you seem to have written + 4*x twice.
 

Flop21

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Re: MATH1131 help thread

How do I do these questions without a calculator?

E.g. Find the argument of -2-2i

Particularly the part where you find tan(theta) = 2/2 or 1, so you must find the inverse of tan to get theta / the angle. How do I get this in a test with no calculator?
 

Flop21

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Re: MATH1131 help thread

For this question:

Determine what conditions on b1,b2,b3,b4 are needed to ensure that <b1,b2,b3,b4> belongs to the span of vectors [I've posted the pic below of it in matrix form]



NOW, these questions are annoying me. I always get different results as there seems to be a number of different ways to solve it or get zeros where I want every time.

I got the answer in the same format as the answers did:



EXCEPT I didn't get the RHS the same.

My RHS was <b1, b2+2b1, b3-2b2, b4+12b3>, thus getting the conditions, b3-2b2 = 0 and b4+12b3 = 0.

But yeah there's no b1 condition there.

Anyway I'm just confused about these questions and what I'm doing wrong. Or are there multiple solutions?
 

InteGrand

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Re: MATH1131 help thread

How do I do these questions without a calculator?

E.g. Find the argument of -2-2i

Particularly the part where you find tan(theta) = 2/2 or 1, so you must find the inverse of tan to get theta / the angle. How do I get this in a test with no calculator?
It is geometrically clear that the (principal) argument is -3pi/4.

I.e. Draw the point (-2,-2) in the x-y plane. Draw the ray from the origin through this point. The angle formed between this and the positive x-axis is clearly 135 degrees (since it lies on the line y = x, which forms a 45-degree line through the origin), or 3pi/4. But it's clockwise instead of counter-clockwise from the positive x-axis, so it's with a negative sign, i.e. the argument is -3pi/4.
 
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InteGrand

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Re: MATH1131 help thread

For this question:

Determine what conditions on b1,b2,b3,b4 are needed to ensure that <b1,b2,b3,b4> belongs to the span of vectors [I've posted the pic below of it in matrix form]



NOW, these questions are annoying me. I always get different results as there seems to be a number of different ways to solve it or get zeros where I want every time.

I got the answer in the same format as the answers did:



EXCEPT I didn't get the RHS the same.

My RHS was <b1, b2+2b1, b3-2b2, b4+12b3>, thus getting the conditions, b3-2b2 = 0 and b4+12b3 = 0.

But yeah there's no b1 condition there.

Anyway I'm just confused about these questions and what I'm doing wrong. Or are there multiple solutions?
There are multiple ways of expressing the conditions required, but they all express the same thing, just in a different way. It's like how expressing a plane in parametric vector form can be done in many different ways, or how a system of simultaneous equations can be expressed equivalently in various ways (these ways being related through elementary row operation transformations).

You can check that your answer isn't equivalent to the solutions' answer by noting that in your one, b1 can be anything at all, completely independent of the other components of the vector b. However, according to the solutions, this is not the case. This means that there are going to be vectors in your answer that won't satisfy the conditions of the solutions' answer, as in your one, we can take b1 to be anything at all regardless of the other three components, whereas if these are specified in the solutions' answer (in some consistent way) then so is b1. So the two answers are not equivalent.

For a specific example, the vector (1,0,0,0) satisfies the conditions you got, but you can check that it doesn't satisfy the solutions' conditions.
 
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Flop21

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Re: MATH1131 help thread

What's going on here? He applies R5-R4, to R5-R1. Why did he get what he did in R5 RHS ?? Especially the R2?



Video not much help since he skips over that step:

https://youtu.be/Cngv6eCOcuQ?t=3m8s
 

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