Help? (1 Viewer)

[airie]

From Cambrudge Year 11 3u textbook, exercise 10F, Q11:
Basically, it involves sketching the graph of x³+y³=a³ where a is a constant, but my graph looks different from that in the answers section. And I can't figure out why

(Note: I'm assuming that a>0 here, same as the textbook. The sign of a is not really the problem here: just reflect the graph for positive a's in the line y=-x for the graph for negative a's.)

Firstly by using implicit differentiation,
x³+y³=a³
3x²+3y².dy/dx=0
dy/dx = -x²/y²

This means that the gradient of the graph is negative for all values of x except at (a,0) where the gradient is undefined, and that the point (0,a) is a stationary point of inflexion, as dy/dx does not change sign as the value of x increases through 0.

Furthermore, using implicit differentiation again to get d²y/dx²,
From dy/dx = -x²/y²,
d²y/dx²=-(2xy²-2x²y.dy/dx)/y⁴
= -(2xy²+2x⁴/y)/y⁴ as dy/dx = -x²/y²,
= -[2x(x³+y³)]/y⁵
= -2xa³/y⁵ as x³+y³=a³

Therefore in the first quadrant (between the two critical values of dy/dx), where x and y are of the same sign, the graph is concave down, and in the second and fourth quadrant, the graph is concave up. There are no points of inflexion other than (0,a) found above using the first derivative.

The graph that I got as the result of my arguments above differs greatly from that shown in the answers section of the textbook, where it is drawn as almost a square but with very rounded corners. The diagram attached shows the graphs in the answers section and as I drew it.

So is the textbook wrong on this? Or am I missing something?

 

[Riviet]

Hmm... Cambridge solutions are RARELY wrong. Your reasoning seems valid at a first glance. The only thing I looked at are the intercepts, I think the Cambridge solutions are assuming that "a" can be both positive and negative.

 

[airie]

o.0 BOTH negative and positive? Wouldn't that be two graphs then?

 

[Affinity]

Are you sure it's not |x|^3 + |y|^3 = a^3 ?

 

[airie]

^Nope, it's just x^3 + y^3 = a^3, no absolute values.

So what went wrong?

 

[fishy89sg]

obviously, chucking that graph into a graphics machine you get your graph:

View attachment 13814

 

[SoulSearcher]

Yeah, but then you don't get a graphics machine during the HSC exam, do you?

 

[fishy89sg]

do i? o.o

 

[Riviet]

do i? o.o

Nope.

I reckon either the question is wrong or the answers are wrong.

 

[SoulSearcher]

I reckon either the question is wrong or the answers are wrong.

Flip a coin

 

[airie]

Right. XD Either way, so the textbook's question and answer don't match each other?

 

[SoulSearcher]

Not unless they somehow take the values of a as both positive and negative, but then that just confuses things even further. It's safe to say that they don't match.

 

[hyparzero]

How interesting.. reminds me of Fermat's last theorem:

xⁿ + yⁿ = aⁿ

 

[airie]

^... That's...kind of...random.

They don't have to be integers, after all *shifty eyes*

 

[SoulSearcher]

Wouldn't work if they were integers

 

[airie]

So it looks like the answer's wrong, after all?

Btw, what do you guys think the equation for the graph drawn in the textbook is? I think it might be x^4+y^4=a^4 for a constant a, but I'm not sure The circle should look more like a square on the corners because when x changes, say increases, y would need to decrease more rapidly due to the higher power of the variables, as opposed to the circle in x^2+y^2=a^2.

 

[SoulSearcher]

Well it's not x⁴+y⁴ = a⁴

 

[airie]

^So what would x^4+y^4=a^4 look like, if not a square-looking circle? ... Maybe I'm making a mistake somewhere?

 

[Riviet]

I shall investigate the original problem posed by airie in this thread.

 

[pLuvia]

I entered it into a graphing software, and it comes out as the Cambridge graph