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C1 - Differentiation and circles

Question 8 ii)

I was just wondering why cant you put y = 3x + 4

in to the equation

( x + 3 ) ^2 + ( y- 2 ) ^2 = 17^2

I did this and got ( x + 3 )^2 + (3x + 2 )^2 = 17^2

took the sqrt of both sides and ended up with x = 3 which is wrong...

Also the 11iii)

The first part x = 5/2

the second part x lies between .... -11/3 x < 3

Final answer = 5/2 less than or equal to x < 3

I was just wondering the final answer does not give the complete range of x values as x can 1 or 2. Is this correct?
Reply 1
Avatar for justinawe
justinawe
19
Would be nice if you could perhaps post the question? :tongue:
Reply 2
Avatar for joostan
joostan
13
Original post by IShouldBeRevising_
Question 8 ii)

I was just wondering why cant you put y = 3x + 4

in to the equation

( x + 3 ) ^2 + ( y- 2 ) ^2 = 17^2

I did this and got ( x + 3 )^2 + (3x + 2 )^2 = 17^2

took the sqrt of both sides and ended up with x = 3 which is wrong...

Square rooting both sides is horribly messy, far better to expand and solve the quadratic.
Reply 3
Avatar for justinawe
justinawe
19
Original post by IShouldBeRevising_
Question 8 ii)

I was just wondering why cant you put y = 3x + 4

in to the equation

( x + 3 ) ^2 + ( y- 2 ) ^2 = 17^2

I did this and got ( x + 3 )^2 + (3x + 2 )^2 = 17^2

took the sqrt of both sides and ended up with x = 3 which is wrong...

Also the 11iii)

The first part x = 5/2

the second part x lies between .... -11/3 x < 3

Final answer = 5/2 less than or equal to x < 3

I was just wondering the final answer does not give the complete range of x values as x can 1 or 2. Is this correct?


One thing I did notice with your working even though I don't know the question is that you've made quite a misconception here.

You did (x+3)2+(y−2)2=x+3+y−2\sqrt{( x + 3 )^2 + ( y- 2 )^2} = x+3+y-2 which is incorrect

Please note that a2+b2≠a+b\sqrt{a^2 + b^2} \not= a+b. Expand out the brackets instead.
(edited 11 years ago)
Original post by justinawe
One thing I did notice with your working even though I don't know the question is that you've made quite a misconception here.

You did (x+3)2+(y−2)2=x+3+y−2\sqrt{( x + 3 )^2 + ( y- 2 )^2} = x+3+y-2 which is incorrect

Please note that a2+b2≠a+b\sqrt{a^2 + b^2} \not= a+b. Expand out the brackets instead.


Yeah I thought they were equal

Original post by joostan
Square rooting both sides is horribly messy, far better to expand and solve the quadratic.




Original post by justinawe
Would be nice if you could perhaps post the question? :tongue:


I always forget to add the question..
http://www.ocr.org.uk/Images/57769-question-paper-unit-4721-01-core-mathematics-1.pdf
Reply 5
Avatar for joostan
joostan
13
Original post by IShouldBeRevising_
Yeah I thought they were equal

I always forget to add the question..
http://www.ocr.org.uk/Images/57769-question-paper-unit-4721-01-core-mathematics-1.pdf


This appears to be the wrong paper no q11 and q8 is different :tongue:
Original post by joostan
This appears to be the wrong paper no q11 and q8 is different :tongue:



:banghead:

http://www.ocr.org.uk/Images/59551-question-paper-unit-4721-01-core-mathematics-1.pdf
Reply 7
Avatar for joostan
joostan
13
Original post by IShouldBeRevising_

For q8 do as we suggested above.
As for q11 part i) is to find the perimeter in terms of x. You appear to have solved for x. . .
EDIT: I see what you mean now just a mo...
No x cannot equal 1 or 2
(edited 11 years ago)
Reply 8
Avatar for justinawe
justinawe
19
Original post by IShouldBeRevising_


For 11iii) your answer is correct and gives the complete range of x values that satisfy the both inequalities. x cannot be 1 or 2 as the perimeter would not be at least 39 m if that were the case.
Original post by justinawe
For 11iii) your answer is correct and gives the complete range of x values that satisfy the both inequalities. x cannot be 1 or 2 as the perimeter would not be at least 39 m if that were the case.



Original post by joostan
For q8 do as we suggested above.
As for q11 part i) is to find the perimeter in terms of x. You appear to have solved for x. . .
EDIT: I see what you mean now just a mo...
No x cannot equal 1 or 2


Ok thank you guys I get it now :smile:
Reply 10
Avatar for joostan
joostan
13
Original post by IShouldBeRevising_
Ok thank you guys I get it now :smile:


Good :smile:

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