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

**0**

# 2002 past paper calc

Started by Dave, May 08 2004 04:19 PM

6 replies to this topic

### #1

Posted 08 May 2004 - 04:19 PM

If i am not here i am somewhere else

### #2

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

If i am not here i am somewhere else

### #3

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).

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

### #4

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

yeah we know we are dumb

### #5

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 -

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 -

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

y = root(3)x + 1 -

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

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

Posted 12 May 2004 - 11:17 PM

Ta, fixed.

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