P2 equation help

this is question 5 from p2 november 2002 and im really stuck can anyone help?
5. The curve C with equation y = p + qex, where p and q are constants, passes through the point (0, 2). At the point P(ln 2, p + 2q) on C, the gradient is 5.(a) Find the value of p and the value of q.

The normal to C at P crosses the x-axis at L and the y-axis at M.
(b) Show that the area of OLM, where O is the origin, is approximately 53.8.

Reply 1

Jonny W

8

(a)
dy/dx = q e^x

At P, dy/dx = 2q. So q = 5/2.

Since the curve passes through (0, 2), p + q = 2. So p = -1/2

(b)
P = (ln 2, 9/2)

The normal at P has gradient -1/5 and equation y = -(1/5)(x - ln(2)) + 9/2. So

L
= ((9/2)/(1/5) + ln(2), 0)
= (23.1931, 0)

M
= (0, (1/5)ln(2) + 9/2)
= (0, 4.63863)

Area(OLM)
= (1/2)*23.1931*4.63863
= 53.8

OP

thank u

this is question 5 from p2 november 2002 and im really stuck can anyone help?
5. The curve C with equation y = p + qex, where p and q are constants, passes through the point (0, 2). At the point P(ln 2, p + 2q) on C, the gradient is 5.(a) Find the value of p and the value of q.

for (0,2)
2 = p + q

dy/dx = qe^x
at p: 5 = q.e^ln2 = 2q
q = 5/2

2 = p + 5/2
p = -1/2

:. y = -1/2 +5/2 e^x

The normal to C at P crosses the x-axis at L and the y-axis at M.
(b) Show that the area of OLM, where O is the origin, is approximately 53.8.

P(ln 2, -1/2 + 2(5/2)) = (ln2 , 9/2)
normal: gradient = -1/5
[y-(9/2)] = -1/5 [x-ln2]

for L , at x-axis means y=0
[0-(9/2)] = -1/5 [x-ln2]
(9)(5)/2 = x-ln2
x = 45/2 + ln2
L(45/2+ln2 , 0 )

for M, at y-axis means x=0
[y-(9/2)] = -1/5 [0-ln2]
y-(9/2) = [ln2]/5
y = [ln2]/5 + 9/2
M(0,[ln2]/5+9/2)

Area of MOL = 1/2 . base . height
= 1/2 . 45/2+ln2 . ([ln2]/5+9/2) = 53.8

edit: why did it take me all that long to post?!?!?! i didnt see ur answer :biggrin:

P2 equation help

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