I can't think of how to label my illustration for this question... I need someone to direct me in the right way.
A baseball diamond is a square with sides of length 90 ft. A batter hits the ball and runs toward first base with a speed of 26 ft/s.
At what rate is his distance from second base changing when he is halfway to first base?

- anonymous

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- anonymous

|dw:1434166696015:dw|

- anonymous

I've misplaced #1 and 3 but it shouldn't make a difference

- anonymous

@ganeshie8 thank god you're here! Can you help me on this?

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## More answers

- ganeshie8

|dw:1434167956990:dw|

- anonymous

Mhmm and how do I label the info given in the question

- ganeshie8

|dw:1434168228668:dw|

- ganeshie8

apply pythagorean theorem, differentiate and do the usual stuff

- wolf1728

Just to make things easier, most people visualize a baseball diamond with home plate at the bottom, 1st base on the left, etc.

##### 1 Attachment

- anonymous

So is this what I have to do...
\[90^2+(90-x)^2=y^2\]
\[2(90-x)=2yy'\]
\[y'=\frac{ 2(90-x) }{ 2y }\]

- Michele_Laino

hint:
using the drawing of @ganeshie8 we can write:
\[y = \sqrt {{{90}^2} + {x^2}} \]
furthermore, please keep in mind that:
\[x = x\left( t \right) = vt = 26t\]

- anonymous

Wouldn't it be
\[y=\sqrt{90^2+(90-x)^2}\]

- Michele_Laino

the right triangle is:
|dw:1434170152802:dw|

- anonymous

I'm a bit confused

- Michele_Laino

when I write:
\[y\left( t \right) = \sqrt {{{90}^2} + x{{\left( t \right)}^2}} \]
am I right?

- Michele_Laino

@Tracy96

- anonymous

yes

- Michele_Laino

|dw:1434170615708:dw|

- Michele_Laino

ok, now try to differentiate y(t) with respect to the time t, what do you get?

- Michele_Laino

please keep in mind that you have to use the chain rule, since:
\[x = x\left( t \right) = vt = 26t\]

- anonymous

Just a quick question, why is it x(t) on one of the legs?

- Michele_Laino

it is the distance between the current position of the player, from the first base

- Michele_Laino

|dw:1434170805543:dw|

- anonymous

Ok got it

- Michele_Laino

ok! Now what is:
\[\frac{{dy}}{{dt}} = y' = ...\]?

- anonymous

Is it
\[y'=\frac{ 1 }{ 2 }\sqrt{(x(t))^2 + 90^2} \times2x(t)x'(t)\] ??

- Michele_Laino

use this rule:
\[y = \sqrt x , \Rightarrow y' = \frac{1}{{2\sqrt x }}\]

- anonymous

\[\frac{ 2x(t)x'(t) }{ 2\sqrt{x(t)^2+90^2} }\]
like that?

- Michele_Laino

that's right!

- anonymous

Relief! haha ok now what?

- Michele_Laino

now, as I wrote before, we have:
\[x = x\left( t \right) = vt = 26t\]

- Michele_Laino

so, what is:
\[x' = \frac{{dx}}{{dt}} = ...?\]

- anonymous

26 ?

- Michele_Laino

perfect!

- Michele_Laino

next, we have to substitute the expression of x and its derivative, into the above firtst derivative of y(t)

- anonymous

Wait so why is x(t) = vt ?
Is that coming from ---> distance = velocity * time

- Michele_Laino

yes! I think that in your exercise, it is supposed that the motion of the player is an uniform motion

- anonymous

Ohh gotcha!
So we first find y' and then x' and then substitute x' into the equation we got from y'... am i following?

- Michele_Laino

yes!

- anonymous

Alrighty! So then
\[y'=\frac{ 52x(t) }{ 2\sqrt{x(t)^2 +90^2} }\] right?

- Michele_Laino

ok! Now simplify:
52/2=26 and substitute x= vt

- anonymous

Ohh so \[y'=\frac{ 26(26t) }{ \sqrt{(26t)^2+90^2} }\]

- Michele_Laino

that's right!

- anonymous

Great!

- Michele_Laino

next we have to compute the time at which the player is at midpoint between the first base and the home plate

- Michele_Laino

in other words we have to solve this equation:
\[45 = 26t\]
what is t?

- anonymous

t = 1.73

- Michele_Laino

ok!

- anonymous

or should i leave it as a fraction 45/26 ?

- Michele_Laino

finally, replace t with 1.73 into your last formula for y' (t), and you will get your answer

- Michele_Laino

better is the fraction form

- anonymous

Right cuz then I can cancel things out and at the end i get 25/90

- Michele_Laino

please wait I'm checking your computation

- anonymous

Oh wait i made a mistake

- anonymous

\[\frac{ 25(45) }{ \sqrt{45^2 +90^2} }\]

- Michele_Laino

ok!

- anonymous

Is that correct?

- Michele_Laino

yes!

- anonymous

Ahhh awesome!! I really bugged you didn't I?

- Michele_Laino

It is all right! :)

- anonymous

Thanks so much!! I have one more question about this problem xD

- Michele_Laino

ok! I can help you!

- anonymous

Oh my goodness I love you so much! :D
It's also asking to find at what rate the player's distance from third base is changing at the same moment... would I have to start all over?

- Michele_Laino

yes! I think so!

- anonymous

Oh no! :S

- Michele_Laino

it is necessary, since we have to rewrite the function y(t), after that the procedure is the same as before

- anonymous

|dw:1434173455402:dw|
is this drawing right?

- Michele_Laino

if we pick as origin of our x-axis the first base, then we have:
|dw:1434173776450:dw|
so:
\[y\left( t \right) = \sqrt {{{90}^2} + {{\left\{ {90 - x\left( t \right)} \right\}}^2}} \]

- anonymous

Ohhh so now I get what's happening!

- anonymous

That's gonna be the only difference correct?

- Michele_Laino

correct!

- anonymous

Otherwise I just do the same thing as before

- Michele_Laino

yes! the procedure is the same as before, only the function y(t) is different

- anonymous

I honestly don't know how to thank you!! You're a lifesaver! Many thanks!

- Michele_Laino

:)

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