C3 help!

A curve has the equation y = √(3x+11)

The point P on the curve has x-coordinate 3.

(a) Show that the tangent to the curve at P has the equation 3x - 4√5y + 31=0

Reply 1

BCHL85

You

A curve has the equation y = √(3x+11)

The point P on the curve has x-coordinate 3.

(a) Show that the tangent to the curve at P has the equation 3x - 4√5y + 31=0

At x = 3, y(3) = rt(9+11) = 2rt(5)
dy/dx = 3/2rt(3x+11)
At x = 3, dy/dx(3) = 3/(2.2rt(5)) = 3/4rt(5).
Then write the equation:
y - y(3) = dy/dx(3) x (x - 3)

Reply 2

You

BCHL85

At x = 3, y(3) = rt(9+11) = 2rt(5)
dy/dx = 3/2rt(3x+11)
At x = 3, dy/dx(3) = 3/(2.2rt(5)) = 3/4rt(5).
Then write the equation:
y - y(3) = dy/dx(3) x (x - 3)

I've got the wrong answer..can you please finish it?

Reply 3

Fermat

y(3) = 2rt(5)
dy/dx(3) = 3/4rt(5)
y - y(3) = dy/dx(3) x (x - 3) becomes

y-2√5 = 3/(4√5)*(x - 3)
4y√5 -8√5√5 = 3x - 9
4y√5 -40 = 3x - 9
3x + 31 - 4y√5 = 0
==============

Reply 4

You

Thanks. Can you do this as well please?

Use the identities for sin(A+B) and sin(A-B) to prove that:

sinP + sinQ = 2sin[(P+Q)/2] cos [(P+Q)/2]

Reply 5

Fermat

sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB

adding the two eqns,

sin(A+B) + sin(A-B) = 2sinAcosB

writing,
P = A+B
Q = A-B
gives,
A = ½(P+Q)
B = ½(P-Q)

giving,

sinP + sinQ = 2sin½(P+Q)cos½(P-Q)
===========================

Reply 6

TheWolf

sin (A+B) + sin (A-B)
=sinAcosB + cosAsinB + sinAcosB - cosAsinB
=2sinAcosB (1)

P=A+B
Q=A-B

B=P-A
Q=A-(P-A)
Q=2A-P
(Q+P)/2=A

A=P-B
Q=P-B-B
Q=P-2B
B=(Q-P)/2

B=(P-Q)/2
A=(P+2)/2

Subsititute ^^ into (1)
=2sin[(P+Q)/2] cos [(P+Q)/2]

Reply 7

You

Thanks.

Last one:

Find in terms of pi, the solutions of the equation

sin5x+sinx=0

for x in the interval 0<x<pi

I've got pi/6, pi/3, pi/2, 3pi/2, 6pi/5, pi

Reply 8

Fermat

Using the info from above about sinP and sinQ,

sin5x + sinx = 2sin3xcos2x
therefore,
sin3xcos2x = 0
=> 3x = npi, or 2x = pi/2 + npi
x = npi/3, or x = pi/4 + npi/2
====================

for 0 < x < pi,

x = pi/3
x = pi/4, 3pi/4
===========

Edit: just noticed, x should be less than pi, not 2pi :biggrin:

Reply 9

You

Ta! What happens if you differentiate ln(x/4)?

Reply 10

Fermat

you reciprocate what's inside the brackets, ie giving 1/(x/4), then multiply that by the differential coefft of what's inside the brackets. ie 1/4.
giving,

1/(x/4)*(1/4)
= (4/x)*(1/4)
= 1/x

Reply 11

username9816

You

Ta! What happens if you differentiate ln(x/4)?

Alternatively to Fermat's solution, this can be done by using the chain rule.

Q.) Find d/dx [ln(x/4)]

Let the function we want to differentiate be: y = ln(x/4)
Let u = x/4 ---> du/dx = 1/4
y = lnu ---> dy/du = 1/u
Hence: d/dx [ln(x/4)] = dy/dx = dy/du * du/dx = 1/(4u) = 1/[4(x/4)]
---> d/dx [ln(x/4)] = 1/x

Reply 12

Fermat

Alternatively, to nima's solution, you could note that,

ln(x/4) = lnx - ln4 and ln 4 is a constant which becomes zero when differentiated.

So, differentiating ln(x/4) is just the same as differentiating lnx - which is 1/x.

Reply 13

You

The curve with equation y= x^(5/2)ln(x/4), x>) crosses the x-axis at the point P.

The normal to the curve at P crosses the y-axis at the point Q

(b) Find the area of triangle OPQ where O is the origin.

Anyone?

Reply 14

Fermat

y= x^(5/2)ln(x/4),
so y = 0 at x = 4
i.e. P is the point(4,0)

y' = (5/2)x^(3/2).ln(x/4) + x^(5/2).(1/x) - using the product rule

at x = 4,
y' = 0 + 4^(5/2).(1/4) = 32*(1/4) = 8
the normal to curve has gradient of m = -1/8
m = -1/8
======
For a gradient of -1/8 with the x-coord = 4 then the the y-coord = (1/8)*4 = 2
So, Q is the point Q(0,2)

OPQ is a right-angled triangle with base = 4 and height = 2
Area = ½.base.height = ½.4.2 = 4
Area = 4
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