Mathematics: Post your doubts here!

[bakhita]

Need help with the 6th question. Could somebody please explain this?

 

[Anum96]

How do i solve this?

(x^2 + y^2) = 2(x^2 – y^2)

Co-ordinates of M ( maximum point)

Solve the square. And the parenthesis.

x^4 +2(x^2)(y^2) + y^4 = 2(x^2) – (2y^2)

Differentiate.

Dy/dx ,

4(x^3) + 2[2*x*(y^2) +2*y*(x^2) (dy/dx)] +4(y^3)(dy/dx) = 4x – 4y(dy/dx)

4(x^3) + 4*x*(y^2) + 4*y*(x^2)(dy/dx) +4(y^3)(dy/dx) = 4x – 4y(dy/dx)

Make dy/dx the subject,

Dy/dx[4*y*(x^2) +4*(y^3) - 4y]= 4x – 4*x*(y^2) -4(x^3)

Dy/dx = (4x – 4*x*(y^2) -4*(x^3))/( 4*y*(x^2) +4*(y^3) - 4y)

Equate to zero since to find stationary points we put dy/dx=0

0 = (4x – 4*x*(y^2) - 4*(x^3))/( 4*y*(x^2) +4*(y^3) - 4y)

All 4’s will cancel.

0 = (x – x*(y^2) - (x^3))

0 = x(1 – (y^2) - (x^2))

0 = 1 – (y^2) – (x^2)

-1 = - (y^2) – (x^2)

( x^2) + (y^2) = 1 & (x^2) = 1 – (y^2)

Now use the equation of the curve and plug in he above value.

((x^2) + ( y^2)) = 2((x^2) – ( y^2))

(1)^2 = 2(1 - (y^2) - (y^2))

1 = 2(1 – 2(y^2))

½ = 1 – 2(y^2)

½ - 1 = - 2(y^2)

- 1/2 = - 2y^2

(1/2)/2 = y^2

y^2 = ¼

y = sqrt(1/4)

y = ½


x ^2 = 1 – y^2

x ^2 = 1 – (1/2)^2

x ^2 = 1 - ¼

x ^2 = ¾

x= sqrt(3/4)

x= (sqrt3)/2 Ans.

 

[Anum96]

Need help with the 6th question. Could somebody please explain this?View attachment 57703

What exactly are vertical and horizontal elements?
Limits?

 

[bakhita]

What exactly are vertical and horizontal elements?
Limits?

...you can't answer me if you don't know
It's part of Integration concept...horizontal element is a strip going horizontally, the area of which we have gotta find using integration, same goes for vertical elements. The main thing is we can find the area under a curve on a graph, by finding integrals of the equation of curve.
If you've covered integration already, you'd know!

 

[Anum96]

...you can't answer me if you don't know
It's part of Integration concept...horizontal element is a strip going horizontally, the area of which we have gotta find using integration, same goes for vertical elements. The main thing is we can find the area under a curve on a graph, by finding integrals of the equation of curve.
If you've covered integration already, you'd know!

I know everything about integration but ive never seen 'vertical or horizontal elements' in any of the questions
My bad
I tried solving your question. Using u substitution. I got the integral but what next? Couldnt figure out the limits so yeahhh

 

[Anum96]

...you can't answer me if you don't know
It's part of Integration concept...horizontal element is a strip going horizontally, the area of which we have gotta find using integration, same goes for vertical elements. The main thing is we can find the area under a curve on a graph, by finding integrals of the equation of curve.
If you've covered integration already, you'd know!

can u tell me the answer.?
im getting 42.67 for the first part. and 0.5 for the second

 

[bakhita]

can u tell me the answer.?
im getting 42.67 for the first part. and 0.5 for the second

Accurate! very nice indeed. You got it right for the first part! but not for the second though, it's the same as first both parts have answer 42.67

 

[bakhita]

can u tell me the answer.?
im getting 42.67 for the first part. and 0.5 for the second

so how did you do it? could you please show me the procedure and explain a little?

 

[Anum96]

Accurate! very nice indeed. You got it right for the first part! but not for the second though, it's the same as first both parts have answer 42.67

oh yes ! *2 sec dance party*

 

[Anum96]

so how did you do it? could you please show me the procedure and explain a little?

sure. give me 5 minutes to type it

 

[bakhita]

oh yes ! *2 sec dance party*

lettme join too!

 

[Anum96]

so how did you do it? could you please show me the procedure and explain a little?

Your concepts are clear. No doubt. Since you know integration. In this question you are not given the limits but youre told that you have to find the area in the first quadrant. This suggests that there is no negative limit. So if you plug in a few values of x in the equation you will get x as 4 when y equals 0. and when x is 0 you get y as 16. Therefore your limits are 4 and 0.
Now just simply integrate the equation. You will get 16x - x^3/3. Plug in the limits and youll get 42.67 I hope that explains for you

 

[Anum96]

lettme join too!

*joins*
That was funnnn. No?

 

[Anum96]

so how did you do it? could you please show me the procedure and explain a little?

You can even make a quick sketch of the graphy. Thats what I did. Then it was clear that 4 and 0 are the required limits

 

[bakhita]

Your concepts are clear. No doubt. Since you know integration. In this question you are not given the limits but youre told that you have to find the area in the first quadrant. This suggests that there is no negative limit. So if you plug in a few values of x in the equation you will get x as 4 when y equals 0. and when x is 0 you get y as 16. Therefore your limits are 4 and 0.
Now just simply integrate the equation. You will get 16x - x^3/3. Plug in the limits and youll get 42.67 I hope that explain for you

yah...I totally get it! ...thank you so much
Actually, I did make a sketch, and found one limit as 4 and since the other (-4) wasn't in the first quadrant, I was kinda confused, but now I get it!

 

[bakhita]

*joins*
That was funnnn. No?

Epic fun!

 

[Anum96]

yah...I totally get it! ...thank you so much
Actually, I did make a sketch, and found one limit as 4 and since the other (-4) wasn't in the first quadrant, I was kinda confused, but now I get it!

Anytimeee

 

[Anum96]

Epic fun!

Hahaha!

 

[bakhita]

Hahaha!

Are you A'Level student?

 

[Anum96]

Are you A'Level student?

Was Gave my papers this november
Wbu?