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# Because C3 is so much fun...(help appreciated) watch

1. so I'm having trouble with a question in the Edexcel book.

The Curve C has equation y = x^5 - 1. The tangent to C at the point C (-1,-2) meets the curve again at the point Q, whose x-coordinate is K.

(a) Show that k is the root of the equation (x^5 - 5x - 4 = 0).

Seeing as the exam is only a couple days away and I need 90% above or else I am quite screwed I can give rep if you're into that sort of thing.

2. (Original post by Dreizhen)
so I'm having trouble with a question in the Edexcel book.

The Curve C has equation y = x^5 - 1. The tangent to C at the point C (-1,-2) meets the curve again at the point Q, whose x-coordinate is K.

(a) Show that k is the root of the equation (x^5 - 5x - 4 = 0).

Seeing as the exam is only a couple days away and I need 90% above or else I am quite screwed I can give rep if you're into that sort of thing.

I'm not entirely sure if this is right, as I don't know how many marks it's worth, but could you not find the equation to the tangent and then do simultaneous equations to solve it? If this was the right method, the tangent equation would come out as y=5x + 3
3. Differentiate the curve to find the gradient of the tangent with x coordinate -1. Use the coordinates of the tangent you have, along with the gradient which you have worked out to get the equation for the tangent. Once you have the equation for the tangent, equal it to the curve, getting the equation that the question states.
4. Differentiate it, put -1 into the equation to get the gradient. Use the y - y1 = m(x-x1) to get the equation of tangent. which gives you y = 5x + 3

sub that into the curve and you get the equation.
5. Thank you guys. The reason I couldn't get the answer was because for some reason I put down the gradient of the tangent as 4 instead of 5.
6. i remember this question, it's nice
7. stumbled across another tricky one if you guys don't mind.

For -pi < x < pi Solve cot x = 3sin2x giving your answers to 3 s.f. where appropriate.

This is what I've done:

cosx/sinx = 3(2sinxcosx)
cosx/sinx = 6sinxcosx
cosx = 6sin^2xcosx
cosx-6sin^2xcosx = 0
cosx (1 - 6sin^2x) = 0

Therefore

cos x = 0

and 1 = 6sin^2x
sin^2x = 1/6
sin^x = sqrt(1/6)

(I really should learn to use Latex one of these days)
The answer at the back of the book says +/-0.421, which I got, and +/- 2.72, which I didn't. Any idea what I did wrong? Again, thanks in advance
8. (Original post by Dreizhen)
stumbled across another tricky one if you guys don't mind.

For -pi < x < pi Solve cot x = 3sin2x giving your answers to 3 s.f. where appropriate.

This is what I've done:

cosx/sinx = 3(2sinxcosx)
cosx/sinx = 6sinxcosx
cosx = 6sin^2xcosx
cosx-6sin^2xcosx = 0
cosx (1 - 6sin^2x) = 0

Therefore

cos x = 0

and 1 = 6sin^2x
sin^2x = 1/6
sin^x = sqrt(1/6)

(I really should learn to use Latex one of these days)
The answer at the back of the book says +/-0.421, which I got, and +/- 2.72, which I didn't. Any idea what I did wrong? Again, thanks in advance
http://www.thestudentroom.co.uk/show...7#post23235587
9. Thing is cos^-1(0) is still 1.57 and not 2.72. I'm going to assume that the book got it wrong, then.
10. sinx = (+/-) 1/sqrt6

Answers in degrees are (+/-)24.0948... & (+/-)155.90515...
Multiply both by pi then divide by 180 to get it in radians and it pops out.
11. (Original post by Wesssty)
sinx = (+/-) 1/sqrt6

Answers in degrees are (+/-)24.0948... & (+/-)155.90515...
Multiply both by pi then divide by 180 to get it in radians and it pops out.
Ooooor you could just use radians mode on your calculator to save yourself the hassle of conversion. I got that part of the answer correct (see +/- 0.421) but thanks anyway.
12. (Original post by Dreizhen)
Ooooor you could just use radians mode on your calculator to save yourself the hassle of conversion. I got that part of the answer correct (see +/- 0.421) but thanks anyway.
That would make it too easy wouldn't it
13. (Original post by Wesssty)
That would make it too easy wouldn't it
your mind works in mysterious ways, brother.

Anyway, another awful question:

Given that y = sqrt(1+x^2), show that dy/dx is sqrt(3)/2 when x = sqrt(3)

This is what I've done.

When x = sqrt(3)

But I am unsure of where to go from here.
14. (Original post by Dreizhen)
your mind works in mysterious ways, brother.

Anyway, another awful question:

Given that y = sqrt(1+x^2), show that dy/dx is sqrt(3)/2 when x = sqrt(3)

This is what I've done.

When x = sqrt(3)

But I am unsure of where to go from here.

Is it just my imagination or have you substituted x=sqrt2 rather than sqrt3?
15. (Original post by Manitude)

Is it just my imagination or have you substituted x=sqrt2 rather than sqrt3?
What've you been smoking mate

(...naah, just messing. Edited. Thanks for pointing that out. Still having problems though.)
16. (Original post by Dreizhen)
What've you been smoking mate

(...naah, just messing. Edited. Thanks for pointing that out. Still having problems though.)
Ah right, I thought that could have been a reason why you couldn't get the solution.
17. (Original post by Dreizhen)
your mind works in mysterious ways, brother.

Anyway, another awful question:

Given that y = sqrt(1+x^2), show that dy/dx is sqrt(3)/2 when x = sqrt(3)

This is what I've done.

When x = sqrt(3)

But I am unsure of where to go from here.
dy/dx = x(1+x^2)^-1/2

= x/sqrt(1+x^2)

input sqrt3

sqrt3/sqrt(1+3)

= sqrt3/2

Voila!

Basically, the dy/dx of yours isnt quite right.
18. (Original post by Dreizhen)
your mind works in mysterious ways, brother.

Anyway, another awful question:

Given that y = sqrt(1+x^2), show that dy/dx is sqrt(3)/2 when x = sqrt(3)

This is what I've done.

When x = sqrt(3)

But I am unsure of where to go from here.
Here's the correct method and answer for you:

I hope this helps.

-Darkfire
Attached Images

19. yup u forgot ot multiply the power by the derivative in the bracket which is 1/2 multiply by x^2

which is x overall
20. Welp I forgot the 2x.
Thanks for pointing out my stupidity guys.

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