# Curve + tangent

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• Jun 25th 2010, 06:34 AM
maggiec
Curve + tangent
Quick question

I was doing this question
Attachment 17992

For a) I got y = pi/2

but to construct a successful argument to prove that the tangent touches C at infinitely many points, how could I do that? Part b) gets you 3 marks in this paper but what are they awarded for?

Thanks
• Jun 25th 2010, 07:19 AM
Ackbeet
I don't think your equation of the tangent is correct. What did you get for the derivative of y?
• Jun 25th 2010, 07:27 AM
maggiec
Quote:

Originally Posted by Ackbeet
I don't think your equation of the tangent is correct. What did you get for the derivative of y?

I got

dy/dx = cosx.sinx.x^(sinx - 1)

when x= pi/2
dy/dx = 0 because cos(pi/2) = 0

and then the tangent passes through point (pi/2, pi/2)

what do you think?
• Jun 25th 2010, 07:29 AM
Ackbeet
Hmm. I don't think your derivative is correct. What techniques are you using to differentiate?
• Jun 25th 2010, 07:32 AM
maggiec
Quote:

Originally Posted by Ackbeet
Hmm. I don't think your derivative is correct. What techniques are you using to differentiate?

y = x^sinx

let u = sinx
then du/dx = cosx

and y = x^u
thus dy/du = u.x^(u-1)

==> dy/dx = dy/du.du/dx
= sinx.x^(sinx-1).cosx
• Jun 25th 2010, 07:39 AM
Ackbeet
Ok, here's the problem. Your differentiation of $\frac{dy}{du}=u x^{u-1}$ is only valid if $u$ is a constant. In this case, it's not. I would recommend logarithmic differentiation.
• Jun 25th 2010, 07:55 AM
maggiec
Quote:

Originally Posted by Ackbeet
Ok, here's the problem. Your differentiation of $\frac{dy}{du}=u x^{u-1}$ is only valid if $u$ is a constant. In this case, it's not. I would recommend logarithmic differentiation.

Ohhhh.

Ok so here's what I got this time round:

lny = sinx.lnx

and simplified it to give
dy/dx = y(lnxcox + sinx/x)

when x=pi/2, y=pi/2
and dy/dx= 1

Hence the tangent has gradient of 1 and passes through (pi/2, pi/2)
so its equation is: y=x

is it correct this time?

if so, how can we prove b?

Thank you!
• Jun 25th 2010, 08:14 AM
Ackbeet
Everything looks good to me now. Question: does the word "touches" mean touching as in a tangent to the curve, or does it just mean intersect?
• Jun 25th 2010, 08:26 AM
maggiec
Quote:

Originally Posted by Ackbeet
Everything looks good to me now. Question: does the word "touches" mean touching as in a tangent to the curve, or does it just mean intersect?

I believe it means that it is a tangent. If it means that it intersects it usually makes that clear by saying "prove that this tangent intersects C..", I've seen that in other questions
• Jun 25th 2010, 08:30 AM
Ackbeet
Ok, so what ideas do you have for part b?
• Jun 25th 2010, 08:37 AM
maggiec
Quote:

Originally Posted by Ackbeet
Ok, so what ideas do you have for part b?

could you equate the equations to see where they touch? (although equating them would give intersections.. but if they dont intersect then maybe it will give you points as to where they touch?) or maybe use the discriminant?
• Jun 25th 2010, 08:40 AM
Ackbeet
If a line is tangent to a curve, then the line and the curve must have equal x and y values at that point. I would say that'd be a great first step. So, what do you get?
• Jun 25th 2010, 08:46 AM
maggiec
Quote:

Originally Posted by Ackbeet
If a line is tangent to a curve, then the line and the curve must have equal x and y values at that point. I would say that'd be a great first step. So, what do you get?

I just get the solution I already knew =(

x^sinx = x
sinx.lnx = lnx
sinx = 1
x = pi/2

but because without using a graphic calculator, I have no idea what y=x^sinx looks like, I get stuck really quickly..
• Jun 25th 2010, 08:48 AM
Ackbeet
I think your solution is only partial. I agree up to $\sin(x)=1$. But how many x's solve that equation? And where do they solve it?
• Jun 25th 2010, 08:56 AM
maggiec
Quote:

Originally Posted by Ackbeet
I think your solution is only partial. I agree up to $\sin(x)=1$. But how many x's solve that equation? And where do they solve it?

Of course.

So x = pi/2, 5pi/2, 9pi/2... there's no limit to the original question.
what do you mean by 'where' do they solve it?
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