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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can someone explain how to do this problem step by step, please

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We 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: \[d=\sqrt{(15035t)^2+(25t)^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is there any way you could draw it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The draw function on the site is tough

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0also what would plug into the t variable, time, to get rid of the t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so you can't draw the figure?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now 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
 4 years ago
Best ResponseYou've already chosen the best response.0i'll try to draw it...it's going to look brutal though hehe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1318178380389:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so we have to take the derivative of d=((15035t)^2+(25t)^2)^(1/2), then plug in 4 for t, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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=(15035t) y=25t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me know what you get

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0well I got 1.379, can you show your step on how you got that number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, expand them first, then take derivative.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You should get:\[\frac{d/dt(15035t)^2+(25t)^2)}{2\sqrt{(15035t)^2+(25t)^2)}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0After 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
 4 years ago
Best ResponseYou've already chosen the best response.0wait did you use natural log to expand it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No. Just expand as usual and you get a polynomial on top. Take the derivative of this polynomial.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is that rule or something because I haven't seen that before except its look similar to natural log

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like for example ln x=1/x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The numerator should be:\[d/dt(225002(150)(35)t+1225t^2+625t^2)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay...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
 4 years ago
Best ResponseYou've already chosen the best response.0I'm just using the chain rule here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and the normal rule for taking derivatives of roots

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if your using the chain for the numerator should you do outside of derivative first which wil be 2(15035t). Then inside which would be 2(15035t)(35)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You dont need the chain rule for the numerator (although you can use it). expand it first then take derivative.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0consider 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
 4 years ago
Best ResponseYou've already chosen the best response.0gotta go. good luck. hope I earned a medal ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lastly did you use (ab)^(2)=a^22ab+b^2 to expand the numerator?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I still didn't get your answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when 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((15035t)^(2)+(25t)^(2))^(1/2), then I plug in 4 and got 200.99

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d/dt(2250010500t+1225t^2+625t^2)}{2\sqrt{(15035t)^2+(25t)^2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This gives:\[\frac{(010500+3700t)}{2\sqrt{(15035t)^2+(25t)^2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now plug in t=4:\[\frac{1480010500}{2\sqrt{(10)^2+(100)^2}}\]Giving,\[\frac{4300}{200.99}\approx21.4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey esidl could you help me with one more problem?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A 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
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[c^2=x^2+y^2\]But y=constant=1. So,\[c=\sqrt{x^21}\]Differentiating:\[dc/dt=\frac{(2x)(dx/dt)}{\sqrt{x^21}}\]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
 4 years ago
Best ResponseYou've already chosen the best response.0oooops 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
 4 years ago
Best ResponseYou've already chosen the best response.0its ok, so its suppose dx/dt=squrt(63)/16 on the denominator? Also could you draw a figure again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no the answer is still dx/dt=sqrt63/8 I just didn't type the 2 in the denominator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1318192994858:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0rectangle is the boat, circle is the pulley, 1m above the level of the boat. distance is the horizontal line

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0again, dc/dt should have been:\[dc/dt=\frac{2x(dx/dt)}{2\sqrt{x^21}}\]cancel the 2 on top with the one on the bottom. everything else works out. answer is still the one given.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no worries...hope is correct (should be!)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when you take the derivative of inside of square root should be 2x on the denominator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0unless 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^21) in the denominator which would be 2X

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I made an error above; c^2=x^2+y^2=x^2+1 I put c^2=x^21 for some unknown reason lol answer should be sqrt(65)/8
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