The integration is with respect to x, so you treat t as a constant for the integration. The point is that whatever value of t you choose, the integral will define a fixed value as a result, so you have a perfectly good function.

Reply 83

Brister

Original post by davros

Thanks, that clears it up nicely :smile:

Reply 84

Brister

Original post by Daniel Freedman

STEP II, 2004, Question 10

Spoiler

I agree that the work done against friction by P and Q in the last part is 24J, but hasn't P only travelled 5 metres at t = 2? Together, P and Q cover a combined distance of 6 metres regardless, so the answer is 24J.

Reply 85

chaosgod69

Original post by Oh I Really Don't Care

TEP III, Question 5

Spoiler

Should the second bracket of first line be Cos(a+b)?

Reply 86

brianeverit

Original post by chaosgod69

Should the second bracket of first line be Cos(a+b)?

Yes

Reply 87

Rexsun

Original post by brianeverit

2004 STEP I question 6

\text{Let mid-points of }BC,CA,AB \text{ be }D,E,F \text{ respectively}

\text{with usual notation, mid-point of }BC \text{ is } \mathbf{d}= \dfrac{1}{2}(\mathbf{b}+ \mathbf{c})

\text{So vector equation of line }Ad \text{ is } \mathbf{r}= \mathbf{a}+ \lambda \left( \dfrac{1}{2}( \mathbf{b}+ \mathbf{c})- \mathbf{a} \right)=(1- \lambda) \mathbf{a}+ \dfrac{\lambda}{2}( \mathbf{b}+ \mathbf{c})

\text{similarly for }BE \text{ we have } \mathbf{r}=(1-\mu) \mathbf{b}+ \dfrac{\mu}{2}(\mathbf{a}+ \mathbf{c})

\text{So at point of intersection we must have }(1- \lambda) \mathbf{a}+ \dfrac{\lambda}{2}( \mathbf{b}+ \mathbf{c})=(1- \mu) \mathbf{b}+ \dfrac{\mu}{2}( \mathbf{a}+ \mathbf{c})

\text{equating coefficients of } \mathbf{a}, \mathbf{b} \text{ and } \mathbf{c} \text{ we have } (1-\lambda)= \dfrac{\mu}{2},\dfrac{\lambda}{2}=(1- \mu) \text{ and }\dfrac{\lambda}{2}= \dfrac{\mu}{2}

\text{Fron whiuch we obtain } \lambda= \mu = \dfrac{2}{3}

\text{Hence, }AD \text{ and }BE \text{ meet at the point }\left( \dfrac{2}{3}(p_1+p_2+p_3),\dfrac{2}{3}(q_1+q_2+q_3)\right)

CF \text{ wil have equation }\mathbf{r}=\mathbf{c}+ \dfrac{\nu}{2}( \mathbf{a}+ \mathbf{b}- \mathbf{c}) \text{ and taking } \nu= \dfrac{2}{3} \text{ gives the same point}

\text{gradient of }AH= \dfrac{q_2+q_3}{p_2+p_3} \text{ and gradient of }BC= \dfrac{q_3-q_2}{p_3-p_2}

\text{so, if they are perpendicular then }\left( \dfrac{q_2+q_3}{p_2+p_30} \right) \left(\dfrac{q_3-q_2}{p_3-p_2} \right)=-1

\Rightarrow (q_2+q_3)(q_3-q_2)=(p_2+p_3)(p_3-p_2) \Rightarrow q_3^2-q_2^2=p_3^2-p_2^2 \Rightarrow p_2^2+q_2^2=p_3^2+q_3^2

\text {similarly, if }BH \text{ and }AC \text{ are at right angles then }p_1^2+q_1^2=p_3^2+q_3^2

\text{ and if both } p_2^2+q_2^2=p_3^2+q_3^2 \text{ and }p_1^2+q_1^2=p_3^2+q_3^2 \text{ then clearly }p_1^2+q_1^2=p_2^2+q_2^2

\Rightarrow (q_1+q_2)(q_2-q_1)=(p_1+p_2)(p_2-p_1) \Rightarrow \left( \dfrac{q_1+q_2}{p_1+p_2}\right) \left( \dfrac{q_2-q_1}{p_2-p_1} \right)=-1 \Rightarrow CH\text{ is perpendicular to }AB

Just to point out one minor mistake that the coordinates of the point of intersection should be ((p1+p2+p3)/2,(q1+q2+q3)/2)
Can somebody change it in the main text?

Reply 88

brianeverit

Original post by Rexsun

Just to point out one minor mistake that the coordinates of the point of intersection should be ((p1+p2+p3)/2,(q1+q2+q3)/2)
Can somebody change it in the main text?

Should actually be ((p1+p2+p3)/3,(q1+q2+q3)/3)
Changed in original post.

Reply 89

fuzzball909

Original post by brianeverit

2004 STEP II question 13

\text{She gains £1 on next draw if she draws another white and loses £}n \text{ if she draws a red}

\text{so expected gain is }1\times Pr\text{(white drawn) }-n\times Pr\text{(red drawn) }=\dfrac{b-n-r}{b-n}- \dfrac{nr}{b-n}= \dfrac{b-n-r-nr}{b-n}

\text{This is zero if }b-n-r-nr=0 \rightarrow n= \dfrac{b-r}{r+1}

\text{with each subsequent throw, the probability of drawing a red increases}

\text{so expected gain de3creases and hence will become negative}

\text{So to maximise her expectation she should stop on the next throw}

\text{after the value found above i.e. when }n= \text{ the integer part of } \dfrac{b-r}{r+1}+1

\text{When }b=2k \text{ say, and }r=1 \text{ she should stop after drawing }\left[ \dfrac{2k-1}{2}+1 \right]=\left[k+ \dfrac{1}{2} \right]=k= \dfrac{b}{2} \text{ balls}

Pr \text{(she draws }\dfrac{b}{2} \text{ white balls and no red)}= \dfrac{b-1}{b} \times \dfrac{b-2}{b-1} \times \dots \times \dfrac{b- \frac{b}{2}}{b- \frac{b}{2}-1}= \dfrac{1}{2}

\text{all other terms cancelling. So expected gain is £}\dfrac{b}{2} \times \dfrac{1}{2}=\text{£} \dfrac{b}{4}

Unparseable latex formula:

\text{Now suppose }b=2k+1, r=1 \text{ the critical value is now }\left[\dfrac{2k+1-1}{2} +1 \right]=k+1 \text{ i.e. }\left(\dfrac{1}2}(b+1)\right)

Pr \text{(she draws} \dfrac{1}{2}(b+1) \text{ white balls) is }\dfrac{b+1}{b}\times \dots \times \dfrac{b-\frac{1}{2}b-\frac{1}{2}}{b-\frac{1}{2}b+\frac{1}{2}}=\dfrac{\frac{1}{2}b-\frac{1}{2}}{b}=\dfrac{1}{2}-\dfrac{1}{2b}

\text{so excpected winnings in this case are } \left(\dfrac{1}{2}-\dfrac{1}{2b}\right) \left( \dfrac{1}{2}b+ \dfrac{1}{2}\right)=\dfrac{(b-1)(b+1)}{4b}=\dfrac{b^2-1}{4b}

How come for the r = 1 case we do not take into account losses when calculating expectation? Surely at every turn there is a 1/b chance she could lose all her money up to that point?