I managed to get the equations for this question but I need a little advice when it comes to differentiating. Question: As part of a school ironman competition, students have to swim across a river which is 30m wide and then run along the side the river until they reach the finishing ling 200 m along the bank. Bob knows that he can run twice as he can swim. If he begins his swim at point X, at what point along the river Y should he get out of the water in order to finish his run at Z in the shortest possible time?

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I managed to get the equations for this question but I need a little advice when it comes to differentiating. Question: As part of a school ironman competition, students have to swim across a river which is 30m wide and then run along the side the river until they reach the finishing ling 200 m along the bank. Bob knows that he can run twice as he can swim. If he begins his swim at point X, at what point along the river Y should he get out of the water in order to finish his run at Z in the shortest possible time?

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My Answer: I used the formula Velocity= Distance/Time and since I need the shortest possible time I rearranged the equation so T= D/V I obtained the equation below: \[t=\frac{ \sqrt{x ^{2+900}} }{ v } + \frac{ 200-x }{ 2v }\] MY QUESTION: Since we want the best point for Bob to swim up to (x)- we differentiate x with respect to t (time) ..... then what about v- as in i dont differentiate it because i am not after optimum velocity
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I guess what I am trying to ask is, is it possible to differentiate one variable and not the other IN ONE EQUATION, and then find the maximum/minimum by making it equal to zero....

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\(\frac{dT}{dx} = 0 \) is the measure of what you are trying to answer go ahead and solve
so i dont differentiate the v ???
no! v is a constant. it could be 10, or 6 or anything
\[t=\frac{ \sqrt{x^2+900} }{ v } + \frac{ 200-x }{ 2v }\] so should i do quotient rule? or differentiate the x expression and leave the v where it is?
put the v on the LHS!
the answer reads: \[\frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\]
which looks good
oh ok, because it is a constant and in this question it is not the velocity we are after, it is the distance ?
no, you are after t, first and foremost.
but we still need to regard the velocity to obtain the best possible distance.. so we dont eliminate it but sort of influence x?
you are minimising t. wrt x so you set dt/dx = 0
v is just a number, it could be 9 or 774 who cares what it is . i don't !! i'm caling it v.
@marigirl if you simplify the expression you got for dt/dx, you should get \[\Large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{v \sqrt{x^2+900} }\] set dt/dx equal to 0 and solve for x. You'll find that v plays no part in the solution to dt/dx = 0. So this means that the velocity v does not alter the point where it's best to get out of the water (for the shortest time possible)
sorry I meant to say \[\Large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{2v \sqrt{x^2+900} }\]
@marigirl you have done it correctly now just finish it off i hate being talked over so i am off. but good luck!
@IrishBoy123 i might have a question for u . give me a min
okay i think i know where im tripping up,, Differentiating (200-x)(2v)
@jim_thompson5910 When u differentiated the equation- did u differentiate each part then add it both together? please tell me what you got when you differentiated the second term .. im just not getting it \[t=\frac{ \sqrt{x^2+900} }{ v }+\frac{ 200-x }{ 2v}\] \[t'=\frac{ x }{ v \sqrt{x^2+900} } + ???\]
v is a constant, so 2v is also a constant along with 1/(2v) pull out the 1/(2v) to just focus on the 200-x derive 200 - x to get ???
-1/2v but then i applied quotient rule and got the first term ... why is it not working for the second term?
I'm not sure what you mean, but we have this \[\large \frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\] simplify the RHS to get \[\large \frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ {\color{red}{2}}*x }{{\color{red}{2}}v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ 1*{\color{red}{\sqrt{x^2+900}}} }{ 2v*{\color{red}{\sqrt{x^2+900}}}}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ \sqrt{x^2+900} }{ 2v\sqrt{x^2+900}}\] \[\large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{ 2v\sqrt{x^2+900}}\]
i guess what you did was treated the V variable as a constant - but didnt eliminate it ...
you mean apply the quotient rule to (200-x)/(2v) ?
yes... i applied quotient rule to the first term .... it is not working for the second for some reason
let's make f(x) = 200-x g(x) = 2v derive both functions to get f ' (x) = -1 g ' (x) = 0 So using the quotient rule to derive f/g, we get [ f ' (x) * g(x) - g ' (x) * f(x) ]/[ g ^2 ] [ -1 * 2v - 0 * (200-x) ]/[ (2v)^2 ] [ -2v - 0 ]/[ (2v)^2 ] (-2v)/(4v^2) -1/(2v) which is the same if we just pull out the 1/(2v) and ignore it temporarily
omggg i was doing silly things all this time!
that's ok
i KNEW v was a constant but i kept differentiating it incorrectly .. omg i wasted so much paper... thanks jim
it's not a waste if you learned something though, so no worries

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