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marigirl
 one year ago
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?
marigirl
 one year ago
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?

This Question is Closed

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2My 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: \[t=\frac{ \sqrt{x ^{2+900}} }{ v } + \frac{ 200x }{ 2v }\] 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

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2dw:1440103563755:dw

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2I 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....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\(\frac{dT}{dx} = 0 \) is the measure of what you are trying to answer go ahead and solve

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2so i dont differentiate the v ???

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0no! v is a constant. it could be 10, or 6 or anything

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2\[t=\frac{ \sqrt{x^2+900} }{ v } + \frac{ 200x }{ 2v }\] so should i do quotient rule? or differentiate the x expression and leave the v where it is?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0put the v on the LHS!

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2the answer reads: \[\frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }\frac{ 1 }{ 2v}\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2oh ok, because it is a constant and in this question it is not the velocity we are after, it is the distance ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0no, you are after t, first and foremost.

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2but we still need to regard the velocity to obtain the best possible distance.. so we dont eliminate it but sort of influence x?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0you are minimising t. wrt x so you set dt/dx = 0

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0v is just a number, it could be 9 or 774 who cares what it is . i don't !! i'm caling it v.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1@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)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1sorry I meant to say \[\Large \frac{ dt }{ dx }=\frac{ 2x \sqrt{x^2+900} }{2v \sqrt{x^2+900} }\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0@marigirl you have done it correctly now just finish it off i hate being talked over so i am off. but good luck!

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2@IrishBoy123 i might have a question for u . give me a min

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2okay i think i know where im tripping up,, Differentiating (200x)(2v)

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2@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 \[t=\frac{ \sqrt{x^2+900} }{ v }+\frac{ 200x }{ 2v}\] \[t'=\frac{ x }{ v \sqrt{x^2+900} } + ???\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1v is a constant, so 2v is also a constant along with 1/(2v) pull out the 1/(2v) to just focus on the 200x derive 200  x to get ???

marigirl
 one year ago
Best ResponseYou've already chosen the best response.21/2v but then i applied quotient rule and got the first term ... why is it not working for the second term?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'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 \[\large \frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ {\color{red}{2}}*x }{{\color{red}{2}}v \sqrt{x^2+900} }\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }\frac{ 1 }{ 2v}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }\frac{ 1*{\color{red}{\sqrt{x^2+900}}} }{ 2v*{\color{red}{\sqrt{x^2+900}}}}\] \[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }\frac{ \sqrt{x^2+900} }{ 2v\sqrt{x^2+900}}\] \[\large \frac{ dt }{ dx }=\frac{ 2x \sqrt{x^2+900} }{ 2v\sqrt{x^2+900}}\]

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2i guess what you did was treated the V variable as a constant  but didnt eliminate it ...

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you mean apply the quotient rule to (200x)/(2v) ?

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2yes... i applied quotient rule to the first term .... it is not working for the second for some reason

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1let's make f(x) = 200x 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 * (200x) ]/[ (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

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2omggg i was doing silly things all this time!

marigirl
 one year ago
Best ResponseYou've already chosen the best response.2i KNEW v was a constant but i kept differentiating it incorrectly .. omg i wasted so much paper... thanks jim

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1it's not a waste if you learned something though, so no worries
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