Question

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?

Answer 1

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

Answer 2

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....

Answer 3

\(\frac{dT}{dx} = 0 \) is the measure of what you are trying to answer


go ahead and solve

Answer 4

so i dont differentiate the v ???

Answer 5

no!

v is a constant. it could be 10, or 6 or anything

Answer 6

\[t=\frac{ \sqrt{x^2+900} }{ v } + \frac{ 200-x }{ 2v }\]

so should i do quotient rule? or differentiate the x expression and leave the v where it is?

Answer 7

put the v on the LHS!

Answer 8

the answer reads:
\[\frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\]

Answer 9

which looks good

Answer 10

oh ok, because it is a constant and in this question it is not the velocity we are after, it is the distance ?

Answer 11

no, you are after t, first and foremost.

Answer 12

but we still need to regard the velocity to obtain the best possible distance.. so we dont eliminate it but sort of influence x?

Answer 13

you are minimising t. wrt x

so you set dt/dx = 0

Answer 14

v is just a number, it could be 9 or 774

who cares what it is . i don't !! i'm caling it v.

Answer 15

@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)

Answer 16

sorry I meant to say

\[\Large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{2v \sqrt{x^2+900} }\]

Answer 17

@marigirl

you have done it correctly

now just finish it off

i hate being talked over so i am off. but good luck!

Answer 18

@IrishBoy123 i might have a question for u . give me a min

Answer 19

okay i think i know where im tripping up,,
Differentiating (200-x)(2v)

Answer 20

@IrishBoy123

Answer 21

@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 \[t=\frac{ \sqrt{x^2+900} }{ v }+\frac{ 200-x }{ 2v}\]

\[t'=\frac{ x }{ v \sqrt{x^2+900} } + ???\]

Answer 22

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 ???

Answer 23

-1/2v

but then i applied quotient rule and got the first term ... why is it not working for the second term?

Answer 24

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
\[\large \frac{ dt }{ dx }=\frac{ x }{v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\]
\[\large \frac{ dt }{ dx }=\frac{ {\color{red}{2}}*x }{{\color{red}{2}}v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\]
\[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ 1 }{ 2v}\]
\[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ 1*{\color{red}{\sqrt{x^2+900}}} }{ 2v*{\color{red}{\sqrt{x^2+900}}}}\]
\[\large \frac{ dt }{ dx }=\frac{ 2x }{2v \sqrt{x^2+900} }-\frac{ \sqrt{x^2+900} }{ 2v\sqrt{x^2+900}}\]
\[\large \frac{ dt }{ dx }=\frac{ 2x- \sqrt{x^2+900} }{ 2v\sqrt{x^2+900}}\]

Answer 25

i guess what you did was treated the V variable as a constant - but didnt eliminate it ...

Answer 26

you mean apply the quotient rule to (200-x)/(2v) ?

Answer 27

yes... i applied quotient rule to the first term .... it is not working for the second for some reason

Answer 28

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

Answer 29

omggg i was doing silly things all this time!

Answer 30

that's ok

Answer 31

i KNEW v was a constant but i kept differentiating it incorrectly .. omg i wasted so much paper... thanks jim

Answer 32

it's not a waste if you learned something though, so no worries