Axioms

What?

How does this add onto any theory? You did take aspects of excisting theories (very vaguely) and tried to incorporate your own idea I think? I hope you know that most of what you said makes absolutely no sence. All the criticism I have .

What exactly do you want to know? Carbon-Carbon bonds are strong but as for the many compounds I think you may have the wrong idea. Look into intermolecular forces and see if this is what you wanted to know. You should probably look at intramolecular forces aswell. It would explain what you need to know in more detail and should answer your question about carbon-carbon bonds.

If the large number of compounds refers to the large variety of carbon chains that we get, then yes. Carbon carbon bonds are very stable and have orbitals where other atoms can join to form a larger compound. To understand this though you need to look at inter- and intramolecular forces.

Without additional restrictions that is definitely wrong. Take the solutions of "x-i=0" and "(x-3)*(3x-i)=0" for example.

It is pretty trivial to check if your indeed found the roots by just plugging your candidate solution into the original polynomial.

 

I don't understand what you are saying there.

 

 

Five minutes by hand, twenty seconds with Wolfram Alpha.

 

You probably calculated something like (4x^3 + 23x^2 +34x -10) * ( x +3 - i) rather than (4x^3 + 23x^2 +34x -10) / ( x +3 - i). But it seems to me that you are not understanding the reason behind doing this in the first place: Any 3rd degree polynomial can be written in the form p(x)=(x-a)(x-b)(x-c). The idea is to find this form, since then solving p(x)=0 is trivial.

If you have i/3-i what answer do u get? I dont understand how you are doing your division?

"I don't understand what you are saying there." Ok lets say z = a +bi then its conjugate is z*= a-bi. Now in z = 0 then zz* must also be equal to zero. It leaves you with a function a^2 +b^2 without the imaginary. It is still a factor because they stated that Zero: -3+i.

It should be true because if I substitute my answer it is equal to zero.

Well, if your calculations are correct then that should of course be the correct solution. I somehow wonder why you think -3-i was a zero? Is that generally true for polynomials with real-valued coefficients that roots must be in some way be related via complex conjugation?

 

What I meant is that you should have found that (4x^3 + 23x^2 +34x -10) / ( x +3 - i) = 4x^2 + (11+4i)x - (3+i), meaning that you are looking for the solutions of (4x^3 + 23x^2 +34x -10) = [4x^2 + (11+4i)x - (3+i)] * [ x +3 - i ] = 0. A product is zero if one of its factors is zero. The two factors are a first and a 2nd order polynomial, respectively. For both cases you should know how to find the zeros.

I think that the conjugate of one of the factors is always a factor. I could be wrong so that is why I am asking. What is true though is that a factor times is conjugate is equal to zero if and only if one of the factors is equal to zero. I know that the one factor is zero so its conjugate multiplied in must also be zero. This is then simple to do with long division because you get rid of the imaginary.

I think your method will work but it seems like a lot of work. I did the working out for the first term using your method and got 4x^4 + 12x^3 +4i/x^2+6x +10. I assume it simplifies at the end?

I have tried that. I cant get a factor of the equation without making the (x+a) part = 10. I think it makes the statement untrue because x=0 when x = -3+i =0.

If it is true then:

4x+11_______

x+3-i |4x^2+23x +34

4x^2 +12x -4xi

It will have a remainder of 1 -4xi +11i

It will mean that 4x + 11 =10 and x = -1/4

Not sure if it makes sense. Maybe if you elaborate a little bit more?

I got the question out of my linear algebra text book. Forms a section on conjugates, modulus and division of complex numbers.

Does this seem correct?

Find all the zeros of the polynomial function by using the given zero

p(x)= 4x^3 + 23x^2 +34x -10 Zero:-3+i

You say that x-(-3+i) and its conjugate x-(-3-i) are your factors. Multiply them together and get x^2 +6x +10

You use long division with this factor to find 4x - 1 and therefore x= 1/4 is another root?

I'm not sure how to start this question. Can someone please assist in explaining how to approach the problem?

Find all the zeros of the polynomial function by using the given zero

p(x)= 4x^3 + 23x^2 +34x -10 Zero:-3+i

1) What is holding us against the Earth?

2) Why is it able to do so?

3) What interaction is there between matter and space?

4) How is it relevant to what is keeping us from being flung from the Earth?

5) What exactly is gravity?

6) Why is G found is both Einstein and Newtons equations?

Fg=Gm1m2/d^2

These are just a few questions to get the discussion going. I think this topic will clarify a lot of what was said in a different topic called "do objects fall at the same speed? no" or something close to that. The arguments mainly revolved around gravity so here is a chance to explain your understanding of what it actually is.

The question itself requires more definition. Of course air resistance affects the rate of fall. Terminal velocity is a function of the atmosphere we fall through and the surface area of the object it self. This puts an entire different question up. It would be beneficial to determine why objects fall at all. What attracts objects? You seem to believe that "gravity" attracts objects and "gravity" is the controller of all objects with mass. Like mass generates a force all its own. Admittedly there is a force that attracts objects. But mass does not generate that force. We acquire mass initially from the definition [mg] and [ma]. Where mass is conserved. Both [mg] and [ma] are subjective measurements of force as we have defined it. [F=mg] and [F=ma}. Where [mg] and [ma] can be objectively measured, calling them force is universalizing what we feel and because what we feel can be quantified as [mg] and [ma] does not justify a force beyond that which we feel and have set equivalent to objective measurements of "resistance".

 

Axioms wrote> "I agree that [g] is a consequence of location. We know that [g] must change when distance between objects increase. An example will be "weightlessness" in space. The weight of the object changes but not its mass. This is what we learn in high school and is a fundamental aspect in physics."

 

This assumes that the attraction is caused by mass. So the further mass is from another mass the weaker the attraction. But you are treating mass as equivalent to matter. Where mass is a measure of resistance and while we feel an equal and opposite resistance to the force we apply, this does not make the resistance a function of what we feel, namely force. And because we can calculate the objective quantities we feel [mg] and [ma] and call them force does not make the universe a function of the force we feel. Both [mg] and [ma] function within the parameters of least action motion. The mathematics functions best with least action motion. So as long as we have quantities like mass [m] and acceleration [a] and [g] and velocity as [v], all of which function within least action motion quantitatively still goes awry because we treat these quantities as though they represent a force beyond the force we apply. And we use mass as a proportional amount of matter beyond its origin which is in classical planetary surface object action.

 

So when we lift an object we say that the planet applies an equal and opposite force that we feel. Whereas the object we lift has a resistance that we can quantify in units of [mg] anywhere we apply the force we feel. The planet exacts a uniform attraction on non-uniform atoms. We define this action as gravitational force where Einstein dealt with it as a uniform gravitational field, still relying on the force we apply to the resistance we work against, to define that resistance we feel as gravity.

 

[F=mg] can also be written as [F=Nnmgx] where [N]=Avogadros Number and [n] = the number of moles and [mg] = the relative weight of a single atom [x]. This applies to the measure of a single isomer of an element but the principle can be applied to all elements. The action we call force then becomes a function of the resistance of a number of atoms where the attraction is uniform on non-uniform atoms. And what we feel is precisely defined.

 

This is much more general of an answer to the original question but it helps to precisely define the terms we use. Have a good time.

johnreed

I think you missed the point of what I said.

You say that mass is not responsible for gravity. You also say that mass does not generate the force of attraction. If we say that Fg=mg what exactly is generating the force of gravity? There must be a gravitational field but what generates this? We know from Newtons laws that Fg= G*m1m1/d^2. The propotional constant G= 6.678*10-11 is known as the universal gravitational constant. This force tends to pull objects towards each other. If we manipulate these equations we find that g= fg/m = GMm/md^2 = GM/d^2. Let us add values. g = (6.67*10^-11)(5.98*10^24)/(6370*10^3)^2 = 9.83m/s^2. These values are the mass of the Earth and the distance of the center of mass to the surface of Earth. If you change the mass of the object you see that the strength of the gravitational field changes. This illustates that the force of gravity is dependent on the mass of the object. It is easy to think of examples. When we landed on the moon, which has less mass than the Earth, its gravitational field is much weaker than here on Earth. If we go to visit the sun the gravity will crush us. If mass did not aid in determining the strength of gravity then theoretically everything would have the same gravitational pull.

If we think in terms of Einstein he illustrated that mass bends space. If you want to go into an argument about what gravity really is we can. If we deny that mass does not play a fundamental role in determining the force of attraction between objects we are blind.

You say we aquired mass initially from the definition of [mg] and [ma]? How exactly if we already have m in the equation? You cannot have a force if there is no mass? Mass is determined atomically and we can relate a large mass to have a center of mass but this is a different topic.

This is so far off the topic where heavier objects will fall faster than lighter objects on Earth outside a vacuum. The question was very specific and it was defined properly. I don't understand why you have ventured off in this direction. I dont know what you are trying to argue here but picture this situation:

You are sitting in a test. You are given the mass of the paper sheet, cardboard sheet and steel sheet with their dimensions being equal. They state it is not in a vacuum. They tell you it is dropped from 50m above the ground. Which will hit the ground first? Explain why this is the case?

Who has the right answer here? I can prove why the steel sheet hits the ground first with mathematics (I have above in my previous post). Your initial statement suggested that they hit the ground at the same time. This is based on the fact that the gravitational field has equal effects on the objects so it must be true that their speeds are equal? Wrong. You need to consider what the question implys and answer accordingly. If you venture off into trying to define what makes it fall to the ground you have missed the point of the question. It is a practical question not theoretical one.

Anyone know what happened on the 28th regarding the tests?

These posts all reflect our ancient beliefs.

 

We used the balance scale to give us weight for thousands of years and for thousands of years we believed that heavy objects fall faster than lighter objects. So when Galileo showed that all objects fall at the same rate when dropped at the same time from the same height we were surprised and we have remained disturbed for 400 years. The problem revolves around what we think is fundamental. A force that we feel and measure and call gravity. We still think we are comparing weight on the balance scale... and we are... but that is superficial. Weight is subject to location.

 

Where we place a balance scale is immaterial to the function of the balance scale (as long as it is placed in a frame within which it can operate). Wherever we place it the magnitude of the acceleration [g] as a factor of the product weight [mg], will be the same on each pan regardless of the mass magnitudes placed on the pans. We calibrate the machine with empty pans or identical weights. So when we define an object in units of weight [mg], the only quantity we are comparing on the balance scale is the quantity of mass [m], since acceleration [g] is a consequence of location, it allows us to calibrate the scale. This uniform action [g] on each pan enables us to compare non-uniform mass resistance on the balance scale. That's pretty simple isn't it? So you might wonder why I bother to point it out.

 

Since what is called gravitational acceleration [g] is a consequence of location and not a consequence of mass [m] (as our tactile sense informs us) all objects MUST fall at the rate of [g]. We should investigate why this is true [1]. We need not waste effort in validating its accuracy. Rather let us determine why it is so.

 

Further, if [g] was not a consequence of location then mass [m] and acceleration [g] could not be combined into the product called weight [mg]. In such a case the balance scale would only give us weight as [w]. However if that were the case we could never have developed.

 

Nonetheless we have engaged in extensive research to verify that all objects fall at the same rate, independent of their mass [m] when dropped at the same time from the same height (in a vacuum). We need not argue that the rate of fall is [g]. Do we need to argue that [g] is a consequence of location? Maybe. If so let us begin. If we agree that [g] is a consequence of location the next question is why.

 

Endnote

[1]

I have made it easier to reference my supporting work by creating a Google Science and Technology Group titled: "The Least Action Consistent Universe and the Mathematics". Currently it contains Sections 1 through 9 for reference. The many sub-sections and work prior to 2007 has not been included. I will develop it further as I have the time and gain familiarity with the venue. Meanwhile my more recent work is available for public review to all, and open to criticism and discussion by any person who joins the group. The latter is a condition established by Google and newsgroups in general. I provide information. I seek no recruits. However, there are no restrictions or requirements to join.

 

Current web address: http://groups.google...oup/thejohnreed

johnreed

I did not deny that gravity effects all objects on earth at the same rate. You used the term in a vacuum where there is no drag. I did specify that I am working outside a vacuum. If I were working in a vacuum then they would all have exactly the same rate of decent of around 9.8m/s^2. There would be no force to oppose the force of gravity.

The main question asked if objects fell at the same speed when outside a vacuum. [g] is the gravitational acceleration that an object experiences. Speed is related to velocity, although [g] still acts the same on both objects, the objects speeds will differ. If this were not true then somebody jumping out of a plane with a parachute would meet a terrible fate. A lot of research has gone into the study of drag which does rely on [g].

I agree that [g] is a consequence of location. We know that [g] must change when distance between objects increase. An example will be "weightlessness" in space. The weight of the object changes but not its mass. This is what we learn in high school and is a fundamental aspect in physics.

A vacuum scenario only illustrates a part of physics which proves that objects experience the same gravitational acceleration. It has been further developed to explain why we can jump out of a plane with a parachute and reach the ground with a constant terminal velocity that we can survive. If you disagree with this then take a peice of cardboard and paper with the same dimentions and drop them. I did say that SA of the objects is a main component when determining drag. When determining how much force needs to be applied by air resistance to let the objects reach terminal velocity is dependent on its mass.

We deal with more than the force of gravity on Earth. There are many forces that we interact with and we need to take this into consideration. If we look at a specific case of falling in a vacuum it is impractical. If you can prove to me that mass does not play a role in determining the speed of an object that is dropped outside of a vacuum I will be impressed.

Please note that I did not deny [g] but am mearly stating that Fg=mg is the amount of force needed to act against the falling object upwards to cause it to reach terminal velocity. I stand by my comment that heavier objects will fall faster than lighter objects when comparing their terminal velocities. If you compare the amount of force needed to make a heavier object reach terminal velocity to that of a lighter object there will be a big difference. The force, SA, air density and drag coefficients will determine how long it would take an object to reach its terminal velocity.

I did specify that in a vacuum they will have the same acceleration and velocity. I did not write all of this for a vacuum, I include air resistance. The study of aerodynamics would be meaninless if we only considered objects falling in a vacuum.

" What we see is the reflection of light that is absorbed by our theoretical view of atoms. "

What?

Was that meant to mean something?

My theoretical view of atoms is a very abstract concept. It doesn't absorb light.

Any light that was absorbed by it would be destroyed by that absorption- it wouldn't be reflected.

Having been absorbed it wouldn't be there to see.

It is supposed to mean something. It was in reply to a quote about how we see bonded atoms. It explains the physical aspects of touch but not why we can see it. Out of context it doesn't make as much sense. I should have explained it in more detail or just worded it properly so I do appologise. If we were to see the atoms bonded with out a light source we would see empty spaces. Light "fills"these empty spaces.

If compounds did not absorb light it would be a "very reflective" world. For an object to appear to have no reflection, or very little reflection, the incoming light source frequency is the same as, or very close to, the vibration frequency of the electrons in the given material. The electrons of the compounds absorb the energy from the light source and thus no reflection.

The light is not destroyed but its photon's energy is absorbed by the electrons in the atom. If you think of the atomic line spectra and supply a high amount of energy to a compound to excite the electrons to a higher energy state, the energy is conserved. When they drop back to their standard state they emit photons, with a frequency, which corresponds to a colour. If this were not true we would find it very difficult to identify compounds.

If you think about it you will see that the colours of everything is determined by its chemical properties. The chemical properties determine what frequencies of light are absorbed and what frequencies of light are reflected.

I am interested to know what your theoretical abstract veiws of atoms are.

I have and it seems to be correct. I dont have a graphing calculator with me though.

The question

r = 1 + sint

Find the points on the cardioid where the tangent line is horizontal or vertical.

I get dy/dt = (cost (1+2sint)) = 0

and dx/dt = (1+sint)(1-2sint) = 0

I can see for dy/dt you get pi/2, 3pi/2, 7pi/2 and 11pi/6

and for dx/dt you get 3pi/2, pi/6 and 5pi/6

I am struggling to remember how to find the r coordinates of this. Do you substitute the points back into the original equation?

You find (2, pi/2) ; etc.

I just need to know if that is how you would get the r coordiante.

If someone can help it would be appreciated.

Sorry, I dont mean question I mean formula.

It is measured in joules and look at this link for how it was derived. E=mc^2 <-- click and go to background. It is summarised but it should be enough.

Ali Algebra know that objects with the same surface area but different densities fall at the same speed in a vacuum.

 

However, there is actually a big argument between me and my girlfriend, at the point of I feel like strangling her to death.

 

I say heavier objects do fall faster (I provided calculations). She (appears) to think that this is nonsense and am full-of-shit and she knows better because she knows more about the topic and am just a uni student (She was my lecturer, before we started dating. Now shes just my bitch)

 

I dropped a piece of paper, and a piece of cardboard with similar SA. The cardboard fell much faster.

 

I could explain this by stating that heavier objects have more force to 'push' air molecules out of the way.

 

In other words, heavier objects are not as affected by air resistance.

I'm not sure if anyone spoke about drag. I just skimmed through the posts. The question did include air resistance though because he stated "know that objects with the same surface area but different densities fall at the same speed in a vacuume" and he did include air resistance in his answer. The paper and cardboard (same shape and different mass) will have different terminal velocities. It is wrong in saying that heavier objects are less affected by air resistance but he has the right idea, kind of.

Terminal velocity is where drag is equal to the force of gravity. Fg_y=ma. It requires less drag force for the lighter object to reach terminal velocity.

proof: v_paper = sqrt(2fg/CpA) (C is the drag coefficient p is the air density and A is the cross-sectional area of the object) (Drag equation manipulated from D=0.5*CpAv^2 and D-Fg=ma)

= sqrt(2(9.8*1*10^-3)/0.5*1*(20*45)) (not sure what A or what p is and cannot be bothered to measure but is meaningless if keep constant)

=6.6*10^-3 m/s

the only thing in the equation that changes is fg so lets say the cardboard is 10* the mass.

v_cardboard = 0.021 m/s

Please note the values I used are not accurate but does not make the calculations wrong. In both equations the only thing that changes is fg so the values are constant in these circumstances.

The objects do accelerate at the same speed initially but the velocity at which they hit the ground will be different. It is dependent on the surface area of the object though. If you take the paper and cardboard and drop it where the maximum SA is expossed and then compare it to where the smallest SA is expossed you will see the difference in results.

Why do replies to this kind of questions always either miss the point entirely or go off on some tangent about stuff we already know?

 

Here's my theory. The OP and his girlfriend are both wrong. The former brings up the empirically correct observation but justifies it poorly. The latter blindly stuck to an elementary fact that was spoon-fed to her and didn't consider that there can be multiple forces acting on an object in this complex universe.

 

Let's state what we know.

 

1) All objects accelerate equally in an ideal vacuum.

2) A paper, cardboard, and steel sheet of the exact same shape have the same surface area.

3) A steel square falls faster than a cardboard square falls faster than a paper square in air.

 

Saying all 3 will fall at the same rate in air is stupid because empirically we can prove this is not true. Saying air resistance as a function of surface area is the only opposing force to gravity is wrong because all 3 have the same surface area. What I feel can only be the proper answer is therefore aerodynamics, which dictates how an object deforms or changes orientation as it is falling, which dictates turbulence of air which affects its rate of fall.

 

A paper sheet bends like crazy in air due to being structurally incapable of maintaining a stable form in the face of pressure from air, thus moving in a sinusoidal pattern and creating some amount of lift in its travel. Its speed is a function of both surface air resistance and lift.

 

A cardboard sheet exhibits similar behavior but to a much lesser extent.

 

A steel sheet exhibits no such behavior, thus it creates no lift due to aerodynamics and its slowdown is purely due to air resistance.

 

Structural integrity is often correlated with mass, so the OP attributed this phenomenon to mass, which makes him wrong. That is to say, if I take a material that is both light and structurally sound - such as a sheet of carbon nanotubes - and dropped it, assuming it did not deform in air, it would fall at the rate more similar to a heavy object of the same shape than a light, malleable object. This is difficult to envision intuitively because such material rarely occurs naturally.

I disagree to a extent. It is related to the mass in the form of force. Lets say the paper, cardboard and steel were all 100% structually sound. We now drop it off a building where all the dimentions of the maximum SA acts towards the ground. The terminal velocity can be determined of each. Drag - Fg =ma. This is the amount of force needed for the air resistance to counteract the force of gravity. Lighter objects will require much less force by air resistance to reach terminal velocity.

Read my last post.

Dean Mullen

Imagine a world were literally nothing existed, something never came into existence and there is just infinite nothing, well you then realized based on maths that even nothing must produce something because:

 

1 divided by 0 = Infinity

thus 0 x Infinity = 1

 

and 0 = nothing & 1 = something thus 0 x Infinity = Infinite nothing which equals to 1 thus

 

eternal nothing = 1 something

 

just as

 

Infinity x 0 = 1

 

so in other words eternal nothing must produce 1/Infinite the size of the nothing of something, thus nothing must create something.

 

Although its hypothetical, if nothing existed in fact, then maybe this equation could be the reason for everything.

Dean Mullen, on 12 February 2011 - 06:40 PM, said:

 

But 1 = 0.9999.... and 1 = 1.0000.......1 so 1 - 1 can equals 1.00000.....1 - 1 = 0.0000....1 so 0.000....1 x Infinity = 1 yet 0.000....1 = 0 so 0 x Infinity = 1 hence if 0 x Infinity = 1 then 1 divided by 0 = Infinity? or is this incorrect.

 

Everything you said here is wrong. You are also basing your whole hypothosis on maths, that we created and not linking it to a physical aspect, to try and explain physical phenomena. There are experiments that show that even in a vaccume something exists by popping in and out of reality. Interesting topics on theories of where these particles come from.

I am busy studying Chemical Engineering.

1) Chemical engineering and nano technology, in the sense I think you want to study, do not meet.

2) Yes, but if you want to get your masters you will need to do research anyways. The research you do will be the main field you will go into. To do research though you need to go into the practical side of the field. Most universities will not let someone with little practical knowledge do their masters or phd. Chemical Engineering is not limited to working in a factory. It has possibly the widest range of working opportunities.

3) There will always be a need for engineers, especially in the years to come.

If you need more information I suggest that you visit a university. You should research what the degree entails and other forms of engineering, if engineering is the direction you want to go into, and see if it suits your personality. This is what I did and I am thoroughly enjoying my course. If you do choose it you will need to enjoy math and chemistry because you do a lot of it.

To answer the original question, we are actually seeing atoms all around us, just that we are not seeing individual atoms, but rather, we see them bonded to one another to form the various compounds we use daily.

The amount of space in atoms is vast. What we see is the reflection of light that is absorbed by our theoretical view of atoms.

akash shrestha

Lepton I meant the similar picture of atom as shown in this link with microscope

http://sciencespot.net/Media/atom2.jpg

In my understanding we will never be able to see the structure described in the picture. In order to see the atom we have to emit light, elecromagnetic radiation, which has energy known as quanta, which the atoms absorb and emits. If we can find a way to look at atoms without projecting light onto it we might see the structure or we might find something completely different. You can look up the atomic line spectra if you want to know more details.

Greippi

Baryon Out of interest, Caesius, what would be needed to "prove" they exist?

We have an idea that atoms behave in a certain manner. We defined a situation where we can measure these ideas which seems to be very appropriate in what we observe. The atom bomb would be an example of the calculations behind the theory of atoms. We cannot prove that atoms exist in the sence of our depictions of the structure of the atom. If we could see electrons in orbit around a nucleus would be an acceptable proof. Unfortunately we cannot do that.

In reality. . .

Ok that makes sense. Thanks makes it simple

Is that not essentially the same thing though?

You still need to define t from the given points and then substitute.

I know to find the slope of parametric equations you find dy/dx.

If you are given the parametric equations and you find an answer in terms of t for the gradient and you are given points in terms of x and y.

E.G.

Find the tangent of

x= 1 + lnt, y= t^2 +2 ; (1,3)

I get dy/dx = 2t^2

Now the text book is not very clear...

What should I do with the 2t^2 and what should I do with this to find the equation of the tangent?

Must I find what t is equal to at that given point or should I eliminate the parameter?

at y = 3

3 = t^2 +2

therefore t = 1 or t = -1 (but t cannot = -1 because of the terms of x for real numbers)

therefore you substitute t = 1 into dy/dx and find the gradient to be 2

I then get y - 3 = 2(x -1)

and get an answer of y=2x +1

I am not sure if its right and if the reasoning is correct or not

If it had the same density as the moon i predict that it would form a black hole.

I guess thats just my logic, the mass will clump together and a massive black hole would form.

Scary thought though

Thanks

Hi I need help to determine if the text books answer is wrong. 6th Edition of Calculus Early Transcendentals by James Stewart if you have the book. Its on page 636 question 3.

It states: Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.

x= t^2 + t , y= t^2 - t ; t = 0

It is basic but the answer they give does not make any sense to me.

They found the equation of the tangent to be (2sint*cost)/(lnt+1)

I found the equation of the tangent to be y=-x

I've drawn the curve and it seems to match my answer at t = 0

If somebody could please confirm my answer or tell me if I'm making some kind of error in my calculation it would be appreciated.