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ta123
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?
Can someone explain how to do this problem step by step, please
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: \[d=\sqrt{(150-35t)^2+(25t)^2}\]
is there any way you could draw it?
The draw function on the site is tough
also what would plug into the t variable, time, to get rid of the t
so you can't draw the figure?
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.
i'll try to draw it...it's going to look brutal though hehe
so we have to take the derivative of d=((150-35t)^2+(25t)^2)^(1/2), then plug in 4 for t, right?
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
let me know what you get
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
well I got 1.379, can you show your step on how you got that number
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.
Yeah, expand them first, then take derivative.
You should get:\[\frac{d/dt(150-35t)^2+(25t)^2)}{2\sqrt{(150-35t)^2+(25t)^2)}}\]
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
wait did you use natural log to expand it?
No. Just expand as usual and you get a polynomial on top. Take the derivative of this polynomial.
is that rule or something because I haven't seen that before except its look similar to natural log
like for example ln x=1/x
The numerator should be:\[d/dt(22500-2(150)(35)t+1225t^2+625t^2)\]
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:\[=\frac{d/dt(w)}{2\sqrt{w}}\]
I'm just using the chain rule here
and the normal rule for taking derivatives of roots
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)
You dont need the chain rule for the numerator (although you can use it). expand it first then take derivative.
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.
gotta go. good luck. hope I earned a medal ;)
lastly did you use (a-b)^(2)=a^2-2ab+b^2 to expand the numerator?
I still didn't get your answer
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
\[\frac{d/dt(22500-10500t+1225t^2+625t^2)}{2\sqrt{(150-35t)^2+(25t)^2}}\]
This gives:\[\frac{(0-10500+3700t)}{2\sqrt{(150-35t)^2+(25t)^2}}\]
Now plug in t=4:\[\frac{14800-10500}{2\sqrt{(10)^2+(100)^2}}\]Giving,\[\frac{4300}{200.99}\approx21.4\]
hey esidl could you help me with one more problem?
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?
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.
\[c^2=x^2+y^2\]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,\[dx/dt=\sqrt{63}/8\]
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
its ok, so its suppose dx/dt=squrt(63)/16 on the denominator? Also could you draw a figure again.
no the answer is still dx/dt=sqrt63/8 I just didn't type the 2 in the denominator
rectangle is the boat, circle is the pulley, 1m above the level of the boat. distance is the horizontal line
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.
no worries...hope is correct (should be!)
when you take the derivative of inside of square root should be 2x on the denominator
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
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