Evaluating the area under a curve

[Krushnaraj Pandya]

Homework Statement

The graph of $y^2 + 2xy + 40 \mid x \mid =400$ divides the plane into regions. The area of the bounded region is?

Homework Equations

All relevant to calculus

The Attempt at a Solution

My question is how to predict the graph without using a grapher here? I put x and y=0 to find the points where it cuts the axes (0,+-20) and (+-10,0). but I never could have guessed what it looks like without a computer (like in my exams)

 

[Mark44]

Homework Statement

The graph of $y^2 + 2xy + 40 \mid x \mid =400$ divides the plane into regions. The area of the bounded region is?

Homework Equations

All relevant to calculus

The Attempt at a Solution

My question is how to predict the graph without using a grapher here?

Split the equation into two parts: one equation in which x < 0 (resulting in $y^2 + 2xy - 40x = 400$) and the other in which x >= 0 (resulting in $y^2 + 2xy + 40x = 400$).

I put x and y=0 to find the points where it cuts the axes (0,+-20) and (+-10,0). but I never could have guessed what it looks like without a computer (like in my exams)

 

[LCKurtz]

Split the equation into two parts: one equation in which x < 0 (resulting in $y^2 + 2xy - 40x = 400$) and the other in which x >= 0 (resulting in $y^2 + 2xy + 40x = 400$).

That problem seems pretty tricky for an in class exam type question. It's easy enough to tell that the graph of each of these is a hyperbola by checking its discriminant. What isn't immediately obvious from either of these equations is that the hyperbolas are degenerate, meaning the graphs are actually the intersecting asymptotes of the would be hyperbolas. And given that they are translated and rotated from the "standard" equations, without a grapher of some sort it would take some significant time to work through the algebra and deduce that the two sections combine to give a parallelogram shaped region.

 

[ehild]

That problem seems pretty tricky for an in class exam type question. It's easy enough to tell that the graph of each of these is a hyperbola by checking its discriminant. What isn't immediately obvious from either of these equations is that the hyperbolas are degenerate, meaning the graphs are actually the intersecting asymptotes of the would be hyperbolas. And given that they are translated and rotated from the "standard" equations, without a grapher of some sort it would take some significant time to work through the algebra and deduce that the two sections combine to give a parallelogram shaped region.

It is easy to see that both expressions $y^2 + 2xy - 40x - 400=0$ and $y^2 + 2xy + 40x - 400=0$ can be written as difference of two squares. $(y^2 + 2xy +x^2) -(x^2\pm40x+400)=0$.
Upon factorizing, any of the factors can be zero, which means four linear functions, enclosing a parallelogram.

 

[Krushnaraj Pandya]

Split the equation into two parts: one equation in which x < 0 (resulting in $y^2 + 2xy - 40x = 400$) and the other in which x >= 0 (resulting in $y^2 + 2xy + 40x = 400$).

I worked that out, its the part that follows which is difficult

 

[Krushnaraj Pandya]

That problem seems pretty tricky for an in class exam type question. It's easy enough to tell that the graph of each of these is a hyperbola by checking its discriminant. What isn't immediately obvious from either of these equations is that the hyperbolas are degenerate, meaning the graphs are actually the intersecting asymptotes of the would be hyperbolas. And given that they are translated and rotated from the "standard" equations, without a grapher of some sort it would take some significant time to work through the algebra and deduce that the two sections combine to give a parallelogram shaped region.

That's high school in India...especially if you want to get into some of the best grad schools here, millions of students pay thousands of dollars to get tuition to learn 2 years worth of an engineering degree while still in the last year of school.
I figured this is a hyperbola, but as you said I need some insight to predict their shape

 

[Krushnaraj Pandya]

It is easy to see that both expressions $y^2 + 2xy - 40x - 400=0$ and $y^2 + 2xy + 40x - 400=0$ can be written as difference of two squares. $(y^2 + 2xy +x^2) -(x^2\pm40x+400)=0$.
Upon factorizing, any of the factors can be zero, which means four linear functions, enclosing a parallelogram.

This is the most helpful insight I think, how did this occur to you so easily? and how can I develop this sort of intuition?

 

[Krushnaraj Pandya]

I think there must be an easier way to evaluate the area using calculus since this problem is given under calculating areas through integrals section of my textbook

 

[ehild]

This is the most helpful insight I think, how did this occur to you so easily? and how can I develop this sort of intuition?

about 70 years practice

 

[Krushnaraj Pandya]

It is easy to see that both expressions $y^2 + 2xy - 40x - 400=0$ and $y^2 + 2xy + 40x - 400=0$ can be written as difference of two squares. $(y^2 + 2xy +x^2) -(x^2\pm40x+400)=0$.
Upon factorizing, any of the factors can be zero, which means four linear functions, enclosing a parallelogram.

Also, how can we figure out either these 4 functions or the vertices of the parallelogram? calculating the area without them isn't possible

 

[Krushnaraj Pandya]

about 70 years practice

hahaha, doesn't seem achievable before my entrance exams.

 

[ehild]

Also, how can we figure out either these 4 functions or the vertices of the parallelogram? calculating the area without them isn't possible

Have you found the two square expressions?
y²+2xy+x² is the square of?
x²±40x+400 is the square of?
How to factorize the difference of two squares ? a²-b²=?

 

[LCKurtz]

hahaha, doesn't seem achievable before my entrance exams.

I guess you had better hurry!
Good catch ehild. I didn't notice that. It's always "easy" after you see it.

 

[Krushnaraj Pandya]

Have you found the two square expressions?
y²+2xy+x² is the square of?
x²±40x+400 is the square of?
How to factorize the difference of two squares ? a²-b²=?

yes I did factorize it but I didn't see that its in the form a^2-b^2 (I'm half asleep right now...)
thank you very much for your help everyone :D
@LCKurtz I'll tell you if I won that race when my exams come, I'll have learned a lot regardless though

 

[ehild]

yes I did factorize it but I didn't see that its in the form a^2-b^2 (I'm half asleep right now...)
thank you very much for your help everyone :D
@LCKurtz I'll tell you if I won that race when my exams come, I'll have learned a lot regardless though

Try to stay awake. You got (x+y)^2-(x±20)^2, did you not? a=x+y, b=x±20. Now factorize, what do you get?
The product is zero, so one factor has to be zero. That means 4 equations, 4 straight lines.

 

[Krushnaraj Pandya]

Try to stay awake. You got (x+y)^2-(x±20)^2, did you not? a=x+y, b=x±20. Now factorize, what do you get?
The product is zero, so one factor has to be zero. That means 4 equations, 4 straight lines.

yes yes, I noted that right away after you hinted at it. I got 2x+y+20, 2x+y-20, y+20 and y-20 and drew the graph right away
:D

 

[Krushnaraj Pandya]

The area is 800 sq. units. Deemed correct by my textbook too.

 

[ehild]

The area is 800 sq. units. Deemed correct by my textbook too.

It was very easy, wasn't it?

 

[Krushnaraj Pandya]

It was very easy, wasn't it?

yes, it really was.
Initially maths was one of my least favorite subjects since I always got stuck on problems like these and the education system here dictates that either the teachers pay attention only to those students who are super-smart and have the ability to score national ranks or do it just as a boring job- considering this, no one had the time or patience to point out simple things such as a^2-b^2 to me.
Now I really enjoy it and am improving, thanks to people like you :D

 

[ehild]

yes, it really was.
Initially maths was one of my least favorite subjects since I always got stuck on problems like these and the education system here dictates that either the teachers pay attention only to those students who are super-smart and have the ability to score national ranks or do it just as a boring job- considering this, no one had the time or patience to point out simple things such as a^2-b^2 to me.
Now I really enjoy it and am improving, thanks to people like you :D

You also could have used the quadratic formula to solve $y^2 + 2xy \pm 40x - 400=0$ equation for y.
Do not forget: Maths is fun! But you need to practice.

 

[Krushnaraj Pandya]

You also could have used the quadratic formula to solve $y^2 + 2xy \pm 40x - 400=0$ equation for y.
Do not forget: Maths is fun! But you need to practice.

did it just to check, got the same answer, thank you :D

 

[ehild]

did it just to check, got the same answer, thank you :D

Well done!

 

[Krushnaraj Pandya]

Well done!

Thank you

 

[LCKurtz]

@Krushnaraj Pandya: Just curious, once you got the parallelogram did you use calculus to find its area or something simpler like vector cross product?

 

[Krushnaraj Pandya]

@Krushnaraj Pandya: Just curious, once you got the parallelogram did you use calculus to find its area or something simpler like vector cross product?

simple 2d geometry, the parallelogram was pretty easy to draw and it had symmetry- besides, vectors is still the next chapter (although I know what a cross product is basically, I don't know how it can be used to calculate area of a 2d figure) and there was no need of calculus