anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1434166696015:dw|
anonymous
  • anonymous
I've misplaced #1 and 3 but it shouldn't make a difference
anonymous
  • anonymous
@ganeshie8 thank god you're here! Can you help me on this?

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More answers

ganeshie8
  • ganeshie8
|dw:1434167956990:dw|
anonymous
  • anonymous
Mhmm and how do I label the info given in the question
ganeshie8
  • ganeshie8
|dw:1434168228668:dw|
ganeshie8
  • ganeshie8
apply pythagorean theorem, differentiate and do the usual stuff
wolf1728
  • wolf1728
Just to make things easier, most people visualize a baseball diamond with home plate at the bottom, 1st base on the left, etc.
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anonymous
  • 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
  • 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
  • anonymous
Wouldn't it be \[y=\sqrt{90^2+(90-x)^2}\]
Michele_Laino
  • Michele_Laino
the right triangle is: |dw:1434170152802:dw|
anonymous
  • anonymous
I'm a bit confused
Michele_Laino
  • Michele_Laino
when I write: \[y\left( t \right) = \sqrt {{{90}^2} + x{{\left( t \right)}^2}} \] am I right?
Michele_Laino
  • Michele_Laino
@Tracy96
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
|dw:1434170615708:dw|
Michele_Laino
  • Michele_Laino
ok, now try to differentiate y(t) with respect to the time t, what do you get?
Michele_Laino
  • Michele_Laino
please keep in mind that you have to use the chain rule, since: \[x = x\left( t \right) = vt = 26t\]
anonymous
  • anonymous
Just a quick question, why is it x(t) on one of the legs?
Michele_Laino
  • Michele_Laino
it is the distance between the current position of the player, from the first base
Michele_Laino
  • Michele_Laino
|dw:1434170805543:dw|
anonymous
  • anonymous
Ok got it
Michele_Laino
  • Michele_Laino
ok! Now what is: \[\frac{{dy}}{{dt}} = y' = ...\]?
anonymous
  • anonymous
Is it \[y'=\frac{ 1 }{ 2 }\sqrt{(x(t))^2 + 90^2} \times2x(t)x'(t)\] ??
Michele_Laino
  • Michele_Laino
use this rule: \[y = \sqrt x , \Rightarrow y' = \frac{1}{{2\sqrt x }}\]
anonymous
  • anonymous
\[\frac{ 2x(t)x'(t) }{ 2\sqrt{x(t)^2+90^2} }\] like that?
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Relief! haha ok now what?
Michele_Laino
  • Michele_Laino
now, as I wrote before, we have: \[x = x\left( t \right) = vt = 26t\]
Michele_Laino
  • Michele_Laino
so, what is: \[x' = \frac{{dx}}{{dt}} = ...?\]
anonymous
  • anonymous
26 ?
Michele_Laino
  • Michele_Laino
perfect!
Michele_Laino
  • Michele_Laino
next, we have to substitute the expression of x and its derivative, into the above firtst derivative of y(t)
anonymous
  • anonymous
Wait so why is x(t) = vt ? Is that coming from ---> distance = velocity * time
Michele_Laino
  • Michele_Laino
yes! I think that in your exercise, it is supposed that the motion of the player is an uniform motion
anonymous
  • 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
  • Michele_Laino
yes!
anonymous
  • anonymous
Alrighty! So then \[y'=\frac{ 52x(t) }{ 2\sqrt{x(t)^2 +90^2} }\] right?
Michele_Laino
  • Michele_Laino
ok! Now simplify: 52/2=26 and substitute x= vt
anonymous
  • anonymous
Ohh so \[y'=\frac{ 26(26t) }{ \sqrt{(26t)^2+90^2} }\]
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
Great!
Michele_Laino
  • 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
  • Michele_Laino
in other words we have to solve this equation: \[45 = 26t\] what is t?
anonymous
  • anonymous
t = 1.73
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
or should i leave it as a fraction 45/26 ?
Michele_Laino
  • Michele_Laino
finally, replace t with 1.73 into your last formula for y' (t), and you will get your answer
Michele_Laino
  • Michele_Laino
better is the fraction form
anonymous
  • anonymous
Right cuz then I can cancel things out and at the end i get 25/90
Michele_Laino
  • Michele_Laino
please wait I'm checking your computation
anonymous
  • anonymous
Oh wait i made a mistake
anonymous
  • anonymous
\[\frac{ 25(45) }{ \sqrt{45^2 +90^2} }\]
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
Is that correct?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
Ahhh awesome!! I really bugged you didn't I?
Michele_Laino
  • Michele_Laino
It is all right! :)
anonymous
  • anonymous
Thanks so much!! I have one more question about this problem xD
Michele_Laino
  • Michele_Laino
ok! I can help you!
anonymous
  • 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
  • Michele_Laino
yes! I think so!
anonymous
  • anonymous
Oh no! :S
Michele_Laino
  • Michele_Laino
it is necessary, since we have to rewrite the function y(t), after that the procedure is the same as before
anonymous
  • anonymous
|dw:1434173455402:dw| is this drawing right?
Michele_Laino
  • 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
  • anonymous
Ohhh so now I get what's happening!
anonymous
  • anonymous
That's gonna be the only difference correct?
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
Otherwise I just do the same thing as before
Michele_Laino
  • Michele_Laino
yes! the procedure is the same as before, only the function y(t) is different
anonymous
  • anonymous
I honestly don't know how to thank you!! You're a lifesaver! Many thanks!
Michele_Laino
  • Michele_Laino
:)

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