Question

Help question 8i!!!! http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_s13_qp_33.pdf

Answer 1

@perl

Answer 2

please show attempt

Answer 3

any attempt at solving :)

Answer 4

tdx2x^2 = k-x^3 dt
2x^2 / k-x^3 dx = dt/t
-2/3 ln(k-x^3) = ln t + c

Answer 5

Then I sub in t=1, x=1 qnd t=4, x=2

Answer 6

ok checking

Answer 7

Then could you pls continue for me ? :(

Answer 8

ok now we sub in

Answer 9

Yeah, but how to solve simultaneously? D:

Answer 10

Brbb.

Answer 11

-2/3 ln(k-x^3) = ln t + c
t=1, x=1

-2/3 ln ( k -1^3 ) = ln(1) + c
-2/3 ln( k - 1) = 0 + c

and t=4, x=2

-2/3 ln( k - 2^3 ) = ln (4) + c
-2/3 ln( k - 8 ) = ln (4) + c

so far we have
-2/3 ln( k - 8 ) = ln (4) + c
-2/3 ln( k - 1) = c

substitute
-2/3 ln( k - 8 ) = ln (4) + -2/3 ln( k - 1)

Answer 12

Ohhh I get it now, thanks,

Answer 13

it is easier if we first simplify the original expression
$$ \Large {
-2/3 \ln(k-x^3) = \ln t + c
\\
\ln [(k-x^3) ^{-2/3}] = \ln (t) + c
\\ \therefore \\
e^{\ln [(k-x^3) ^{-2/3}] } = e^{ \ln (t) + c}
\\ \therefore \\
{(k-x^3) ^{-2/3} } = t \cdot e^c
\\ \therefore \\
\frac{1}{{(k-x^3) ^{2/3} }} =At

}

$$

thats a much easier expression to plug in

Answer 14

wht r u waiting for @perl