anonymous
  • anonymous
At noon, ship A is 150 km west of ship B. ship A is sailing east at 35 km/h and ship B is sailing north at 25km/h. How fast is the distance between the ships changing at 4:00PM?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Can someone explain how to do this problem step by step, please
anonymous
  • anonymous
We need a function describing distance between the ships as a function of the time. The initial distance between the ships is 150km, which occurs at what I will call time t=0. I will call the East direction my positive x and North my positive y. Lets refer everything from the point of view ship A which we will call at rest. Then ship B is approaching A with a horizontal component and moving away from A with a vertical component. Calling d=distance between the ships, we have: \[d=\sqrt{(150-35t)^2+(25t)^2}\]
anonymous
  • anonymous
is there any way you could draw it?

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anonymous
  • anonymous
The draw function on the site is tough
anonymous
  • anonymous
also what would plug into the t variable, time, to get rid of the t
anonymous
  • anonymous
so you can't draw the figure?
anonymous
  • anonymous
Now that we have our function we need to know how fast it changes with time. t=0 is noon here and t=4 would be 4:00pm. We find the derivative of this function (with respect to time) and then plug in t=4 to find out how fast the distance between the ships is changing at 4:00pm.
anonymous
  • anonymous
i'll try to draw it...it's going to look brutal though hehe
anonymous
  • anonymous
|dw:1318178380389:dw|
anonymous
  • anonymous
so we have to take the derivative of d=((150-35t)^2+(25t)^2)^(1/2), then plug in 4 for t, right?
anonymous
  • anonymous
The rectangle on the left is ship A and the rectangle on the right is ship B. d=distance between them (which has contributions from the horizontal distance,x and the vertical distance, y) x=(150-35t) y=25t
anonymous
  • anonymous
Yep
anonymous
  • anonymous
let me know what you get
anonymous
  • anonymous
I get a rate of change of about +21.4km/h at time t=4 (4:00pm). i.e., the distance between the ships is increasing by this amount at 4:00pm
anonymous
  • anonymous
well I got 1.379, can you show your step on how you got that number
anonymous
  • anonymous
maybe I didn't take the derivative right because those squares inside the square root was kinda confusing on how to take the derivative of those properly.
anonymous
  • anonymous
Yeah, expand them first, then take derivative.
anonymous
  • anonymous
You should get:\[\frac{d/dt(150-35t)^2+(25t)^2)}{2\sqrt{(150-35t)^2+(25t)^2)}}\]
anonymous
  • anonymous
After you expand and take the derivative of the numerator you have you function of how the distance changes with time. Plug in t=4 to find the value at 4:00pm
anonymous
  • anonymous
wait did you use natural log to expand it?
anonymous
  • anonymous
No. Just expand as usual and you get a polynomial on top. Take the derivative of this polynomial.
anonymous
  • anonymous
is that rule or something because I haven't seen that before except its look similar to natural log
anonymous
  • anonymous
like for example ln x=1/x
anonymous
  • anonymous
The numerator should be:\[d/dt(22500-2(150)(35)t+1225t^2+625t^2)\]
anonymous
  • anonymous
okay...going back to the distance function\[d=\sqrt{w}\]where w=all the stuff under the root sign Take the derivative with respect to t:\[=\frac{d/dt(w)}{2\sqrt{w}}\]
anonymous
  • anonymous
I'm just using the chain rule here
anonymous
  • anonymous
and the normal rule for taking derivatives of roots
anonymous
  • anonymous
if your using the chain for the numerator should you do outside of derivative first which wil be 2(150-35t). Then inside which would be 2(150-35t)(35)
anonymous
  • anonymous
You dont need the chain rule for the numerator (although you can use it). expand it first then take derivative.
anonymous
  • anonymous
consider the function\[y=\sqrt{x^3}\]To take the derivative you use substitution rule: let u=x^3 \[y=\sqrt{u}\]Now take derivative wrt x:\[dy/dx=(dy/du)(du/dx)=(\frac{1}{2\sqrt{u}})(3x^2)\]No get rid of u, using u=x^3:\[dy/dx=\frac{3x^2}{2\sqrt{x^3}}\]Your problem is analogous to this.
anonymous
  • anonymous
gotta go. good luck. hope I earned a medal ;)
anonymous
  • anonymous
lastly did you use (a-b)^(2)=a^2-2ab+b^2 to expand the numerator?
anonymous
  • anonymous
I still didn't get your answer
anonymous
  • anonymous
when you comeback could you please show the format of your final polynomial function after taking the derivative of the numerator? mine was -10500+3700t/2((150-35t)^(2)+(25t)^(2))^(1/2), then I plug in 4 and got 200.99
anonymous
  • anonymous
\[\frac{d/dt(22500-10500t+1225t^2+625t^2)}{2\sqrt{(150-35t)^2+(25t)^2}}\]
anonymous
  • anonymous
This gives:\[\frac{(0-10500+3700t)}{2\sqrt{(150-35t)^2+(25t)^2}}\]
anonymous
  • anonymous
Now plug in t=4:\[\frac{14800-10500}{2\sqrt{(10)^2+(100)^2}}\]Giving,\[\frac{4300}{200.99}\approx21.4\]
anonymous
  • anonymous
Thanks!
anonymous
  • anonymous
np :)
anonymous
  • anonymous
hey esidl could you help me with one more problem?
anonymous
  • anonymous
sure
anonymous
  • anonymous
A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate 1 m/s, how fast is the boat approaching the dock when it is 8 m from dock?
anonymous
  • anonymous
Ok. We have another triangle here. The hypotenuse is shrinking at a rate of 1 m/s. The height of the triangle is constant at 1m. We want to know how fast the length, x, of the triangle is shrinking when x=8m.
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[c^2=x^2+y^2\]But y=constant=1. So,\[c=\sqrt{x^2-1}\]Differentiating:\[dc/dt=\frac{(2x)(dx/dt)}{\sqrt{x^2-1}}\]Okay, we have a lot of stuf here now. Keeps your knowns and unknowns in mind. dc/dt is known, it is the rate the hypotenuse is shrinking i.e., 1m/s. We want dx/dt. What about x? Well, we want dx/dt when x=8 so this is known. We have:\[1=\frac{8(dx/dt)}{\sqrt{63}}\]So,\[dx/dt=\sqrt{63}/8\]
anonymous
  • anonymous
oooops I made a boo boo. Forgot the factor of 2 in the denominator when I differentiated. But the answer take this into account, just a typo
anonymous
  • anonymous
its ok, so its suppose dx/dt=squrt(63)/16 on the denominator? Also could you draw a figure again.
anonymous
  • anonymous
no the answer is still dx/dt=sqrt63/8 I just didn't type the 2 in the denominator
anonymous
  • anonymous
2X8=18
anonymous
  • anonymous
oops! I meant 16
anonymous
  • anonymous
|dw:1318192994858:dw|
anonymous
  • anonymous
rectangle is the boat, circle is the pulley, 1m above the level of the boat. distance is the horizontal line
anonymous
  • anonymous
again, dc/dt should have been:\[dc/dt=\frac{2x(dx/dt)}{2\sqrt{x^2-1}}\]cancel the 2 on top with the one on the bottom. everything else works out. answer is still the one given.
anonymous
  • anonymous
ok thanks so much
anonymous
  • anonymous
no worries...hope is correct (should be!)
anonymous
  • anonymous
when you take the derivative of inside of square root should be 2x on the denominator
anonymous
  • anonymous
unless that 2 on denominator was suppose to represent the 1/2 and the numerator represented taking the derivative of the inside of the squrt(x^2-1) in the denominator which would be 2X
anonymous
  • anonymous
I made an error above; c^2=x^2+y^2=x^2+1 I put c^2=x^2-1 for some unknown reason lol answer should be sqrt(65)/8
anonymous
  • anonymous
ok, Thanks again

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