C3 June 2014 Q8

Notes

M1 Sub \(t=0\) into \(P\) and use \(\mathrm{e}^0=1\) in at least one of the two cases. Accept \(P=\dfrac{800}{1+3}\) as evidence

A1 200. Accept this for both marks as long as no incorrect working is seen.

Notes

M1 Sub \(P=250\) into \(P=\dfrac{800\mathrm{e}^{0.1t}}{1+3\mathrm{e}^{0.1t}}\), cross multiply, collect terms in \(\mathrm{e}^{0.1t}\) and proceed to \(A\mathrm{e}^{0.1t}=B\)
Condone bracketing issues and slips in arithmetic.
If they divide terms by \(\mathrm{e}^{0.1t}\) you should expect to see \(C\mathrm{e}^{-0.1t}=D\)

A1 \(\mathrm{e}^{0.1t}=5\) or \(\mathrm{e}^{-0.1t}=0.2\)

M1 Dependent upon gaining \(\mathrm{e}^{0.1t}=E\), for taking ln's of both sides and proceeding to \(t=\ldots\)
Accept \(\mathrm{e}^{0.1t}=E\Rightarrow 0.1t=\ln E\Rightarrow t=\ldots\) It could be implied by \(t=\) awrt 16.1

A1 \(t=10\ln(5)\)
Accept exact equivalents of this as long as \(a\) and \(b\) are integers. Eg. \(t=5\ln(25)\) is fine.

Notes

M1 Scored for a full application of the quotient rule and knowing that \(\dfrac{\mathrm{d}}{\mathrm{d}t}\mathrm{e}^{0.1t}=k\mathrm{e}^{0.1t}\) and NOT \(kt\mathrm{e}^{0.1t}\)
If the rule is quoted it must be correct.
It may be implied by their \(u=800\mathrm{e}^{0.1t},v=1+3\mathrm{e}^{0.1t},u'=p\mathrm{e}^{0.1t},v'=q\mathrm{e}^{0.1t}\) followed by \(\dfrac{vu'-uv'}{v^2}\).
If it is neither quoted nor implied only accept expressions of the form \(\dfrac{(1+3\mathrm{e}^{0.1t})\times p\mathrm{e}^{0.1t}-800\mathrm{e}^{0.1t}\times q\mathrm{e}^{0.1t}}{(1+3\mathrm{e}^{0.1t})^2}\)
Condone missing brackets.
You may see the chain or product rule applied to
For applying the product rule see question 1 but still insist on \(\dfrac{\mathrm{d}}{\mathrm{d}t}\mathrm{e}^{0.1t}=k\mathrm{e}^{0.1t}\)
For the chain rule look for
\(P=\dfrac{800\mathrm{e}^{0.1t}}{1+3\mathrm{e}^{0.1t}}=\dfrac{800}{\mathrm{e}^{-0.1t}+3}\Rightarrow\dfrac{\mathrm{d}P}{\mathrm{d}t}=-800\times\left(\mathrm{e}^{-0.1t}+3\right)^{-2}\times-0.1\mathrm{e}^{-0.1t}\) (corrected from the printed mark scheme: the printed line has \(800\times\left(\mathrm{e}^{-0.1t}+3\right)^{-2}\times-0.1\mathrm{e}^{-0.1t}\), without the leading minus sign)

A1 A correct unsimplified answer to \(\dfrac{\mathrm{d}P}{\mathrm{d}t}=\dfrac{(1+3\mathrm{e}^{0.1t})\times 800\times 0.1\mathrm{e}^{0.1t}-800\mathrm{e}^{0.1t}\times 3\times 0.1\mathrm{e}^{0.1t}}{(1+3\mathrm{e}^{0.1t})^2}\)

M1 For substituting \(t=10\) into their \(\dfrac{\mathrm{d}P}{\mathrm{d}t}\), NOT \(P\)
Accept numerical answers for this. 2.59 is the numerical value if \(\dfrac{\mathrm{d}P}{\mathrm{d}t}\) was correct

A1 \(\dfrac{\mathrm{d}P}{\mathrm{d}t}=\dfrac{80\mathrm{e}}{(1+3\mathrm{e})^2}\) or equivalent such as \(\dfrac{\mathrm{d}P}{\mathrm{d}t}=80\mathrm{e}(1+3\mathrm{e})^{-2},\ \dfrac{80\mathrm{e}}{1+6\mathrm{e}+9\mathrm{e}^2}\)

Note that candidates who substitute \(t=10\) before differentiation will score 0 marks

Notes

B1 Accept solutions from substituting P=270 and showing that you get an unsolvable equation
Eg. \(270=\dfrac{800\mathrm{e}^{0.1t}}{1+3\mathrm{e}^{0.1t}}\Rightarrow-27=\mathrm{e}^{0.1t}\Rightarrow 0.1t=\ln(-27)\) which has no answers.
Eg. \(270=\dfrac{800\mathrm{e}^{0.1t}}{1+3\mathrm{e}^{0.1t}}\Rightarrow-27=\mathrm{e}^{0.1t}\Rightarrow\mathrm{e}^{0.1t}/\mathrm{e}^x\) is never negative
Accept solutions where it implies the max value is 266.6 or 267. For example accept sight of \(\dfrac{800}{3}\), with a comment ‘so it cannot reach 270’, or a large value of \(t\) (\(t>99\)) being substituted in to get 266.6 or 267 with a similar statement, or a graph drawn with an asymptote marked at 266.6 or 267
Do not accept exp's cannot be negative or you cannot ln a negative number without numerical evidence.
Look for both a statement and a comment