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2002 past paper calc - HSN forum

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2002 past paper calc


6 replies to this topic

#1 Dave

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Posted 08 May 2004 - 04:19 PM

Could someone help me with question 6

find the eqution of the tangent to the curve y = 2sin(x-pi/6) at the point where x = pi/3

and question 7

Find the x-coordinate of thr point where the graph of the curve with equation y = log base 3 (x-2) +1 intersects the x-axis

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#2 Dave

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Posted 08 May 2004 - 06:03 PM

yeah they were easy enough and i knew the method of doing then but i just wasn't getting the right answer i was maybe too tired

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#3 AndyW

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Posted 08 May 2004 - 06:23 PM

You don't need to use the double angle formule to expand sin (x - pi/6)

This is just of the form sin (ax + n). Differentiating that gives a cos (ax + n).

#4 jeffers

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Posted 11 May 2004 - 07:44 PM

erm, me and my friend still cant do this question, could you please do question 6 with all workings, thanks

yeah we know we are dumb rolleyes.gif

#5 AndyW

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Posted 11 May 2004 - 10:35 PM

(An earlier post on this page has been deleted for some reason... ???)

y = 2 sin (x - pi/6)

Let u = x - pi/6, so y = 2 sin u

Then use the chain rule:

dy/dx = dy/du * du/dx
= 2 cos u * 1
= 2 cos (x - pi/6)

So f'(x) = 2 cos (x - pi/6) and at the point x = pi/3:

f'(pi/3) = 2 cos (pi/3 - pi/6) = 2 cos (pi/6) = root(3)

This is the gradient at the tangent point.

So for the equation of the tangent, we now have the x co-ordinate and gradient, but still need the y co-ordinate. To get this just substitute pi/3 into the original equation:

y = 2 sin(pi/3 - pi/6) = 2 sin (pi/6) = 1

Now just put these into the equation for a straight line:

y - 1 = root3 (x - pi/3)
y = root(3)x + 1 - (root(3)pi)/3

The part is bold is in a different form from that given in the solutions, but is numerically equal. I would think this would still get full marks.

To get into the form in the solutions:

Multiply the top and bottom of the fraction in bold by root(3) to get:

y = root(3)x + 1 - (3pi)/3root(3)

The 3 on the top and bottom then cancel out to give:

y = root(3)x + 1 - pi/root(3)


#6 George

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Posted 12 May 2004 - 05:45 PM

The slip is in the last line, multiplying out the bracket. It should be root(3) times x, not root 3x.

#7 AndyW

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Posted 12 May 2004 - 11:17 PM

Ta, fixed.





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