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Q3C(ii) + Q5B(iii) (1 Viewer)

egnaro

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i cant really be stuffed doing Q3c.ii) with out the right lettering and such so i'll do Q5biii.

To find the inverse (f-¹(x)), you need to change y into x and x into y, hence

original eqn = y = 1/(1+x²)
thus new eqn = x = 1/(1+y²)

which = x(1+y²) = 1
= y² = 1/x - 1
= y = sqrt(1/x -1)
= y= sqrt ((1-x)/x) = inverse of f(x)
 

DAAVE

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3ci:

x^2+h^2 = 16

x^2 = 16-h^2
x = sqrt(16-h^2) [since x is +ve]

3cii:

when h = 1

dh/dt = -.3 (since h is getting smaller)

also:

dx/dh = -1/2 * (16-h^2)^(-1/2) * (-2h)

dx/dt = dh/dt * dx/dh

etc etc... sub in 1 and u get ur answer...
 

bevstarrunner

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5 b iii) replace x with y

x = 1/(1+y²)
1+y² = 1/x
y² = (1/x) - 1
y = + root[(1/x)-1)
 

~ ReNcH ~

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DAAVE said:
3ci:

x^2+h^2 = 16

x^2 = 16-h^2
x = sqrt(16-h^2) [since x is +ve]

3cii:

when h = 1

dh/dt = -.3 (since h is getting smaller)

also:

dx/dh = -1/2 * (16-h^2)^(-1/2) * (-2h)

dx/dt = dh/dt * dx/dh

etc etc... sub in 1 and u get ur answer...
In the end, was your result +ve or -ve?
I got something like 0.08m/hr, but I think I lost a mark because I said "h" was -1, instead of saying dh/dt was -0.3 - I knew something was -ve but I didn't quite know what :)
 

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