Related rates differentiation problem

[frosty8688]

1. At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35 km/hr and ship B is sailing north at 25 km/hr. How fast is the distance between the ships changing at 4:00 pm?



2. None



3. I have the distance as 150 km. I have the variables \frac{dx}{dt} = 35 and \frac{dy}{dt} = 25 and \frac{dZ}{dt} = ? I am just wondering on how to set up the equation.

 

[eumyang]

What's the relationship between x, y, and Z? Note that Z is the hypotenuse of a right triangle.

 

[frosty8688]

x is ship A's position, y is ship B's position and Z is the rate of change in distance.

 

[Vorde]

Realize that the distance equation is the Pythagorean equation, so try setting up a triangle with what you know, and see if you can figure out how to get the hypotenuse of the triangle from what you are given.

 

[frosty8688]

Ok, thanks for the help.

 

[eumyang]

x is ship A's position, y is ship B's position and Z is the rate of change in distance.

Oops, I misread the part that stated that A started west of B and is traveling east. Suppose we call the point where ship B started point O. |AO| = 150. As time goes on, A gets closer to O. What expression would give us the distance |AO| in terms of x?

Also, Z is the distance between A and B. It is not the rate of change in distance.

 

[frosty8688]

The equation would be: \frac{dz}{dt} = \frac{(2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}}{2z}

 

[eumyang]

The equation would be: \frac{dz}{dt} = \frac{(2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt}}{2z}

No. First, the distance |AO| would be 150 - x, not 150 + x. 2nd, show us the equation BEFORE taking the derivative.

 

[frosty8688]

The equation before taking the derivative is (150 - x)^{2} + y^{2} = z^{2}

 

[eumyang]

The equation before taking the derivative is (150 - x)^{2} + y^{2} = z^{2}

I think I see what you did earlier. On the side with the x and y, take the derivative separately, and put the dx/dt and dy/dt next to each term. Earlier, you wrote (2(150 + x) + 2y)\frac{dx}{dt} + \frac{dy}{dt} (ignoring the 150+x error), which makes no sense. The numerator should be written as
(derivative of (150 - x)²)(dx/dt) + (derivative of y²)(dy/dt).

 

[frosty8688]

It would be \frac{dz}{dt} = \frac{(150 - x)\frac{dx}{dt} + y \frac{dy}{dt}}{z}

 

[eumyang]

It would be \frac{dz}{dt} = \frac{(150 - x)\frac{dx}{dt} + y \frac{dy}{dt}}{z}

Nope. You didn't take the derivative of (150 - x)² correctly. Nor the derivative of y².

 

[frosty8688]

The derivative would be \frac{2(150-x)\frac{dx}{dt} + 2y\frac{dy}{dt}}{2z}

 

[eumyang]

The derivative would be \frac{2(150-x)\frac{dx}{dt} + 2y\frac{dy}{dt}}{2z}

Close. You need a negative in front of the 2(150 - x)(dx/dt). (You need to use the chain rule.) Also, to simplify things, factor out the 2 in the numerator and cancel with the 2 in the denominator.

Find the distance ship A traveled (x) in 4 hours. Find the distance ship B traveled (y) in 4 hours. Find Z. Then plug everything in the dZ/dt equation.

 

[frosty8688]

What I got is \frac{4300}{10\sqrt{101}} km/hr

 

[frosty8688]

Which is about 42.79 km/hr

 

[eumyang]

What I got is \frac{4300}{10\sqrt{101}} km/hr

That's not what I got. Can you please show all of your work here, instead of just writing the final answer?

 

[frosty8688]

35 * 4 = 140 for x, 25 * 4 = 100 for y; \frac{-2(150-140)*35+2*100*25}{\sqrt{(150-140)^{2}+ 100^{2}}} = \frac{-2*10*35 + 5000}{\sqrt{10^{2} + 100^{2}}} = \frac{-700 + 5000}{\sqrt{10100}} = \frac{4300}{10\sqrt{101}}

 

[eumyang]

35 * 4 = 140 for x, 25 * 4 = 100 for y; \frac{-2(150-140)*35+2*100*25}{\sqrt{(150-140)^{2}+ 100^{2}}} = \frac{-2*10*35 + 5000}{\sqrt{10^{2} + 100^{2}}} = \frac{-700 + 5000}{\sqrt{10100}} = \frac{4300}{10\sqrt{101}}

Looks like you removed the 2 in the denominator, but never factored out or canceled the 2's in the numerator. Let's just forget about canceling out the 2. You should have written:
\frac{-2(150-140) \cdot 35+2 \cdot 100 \cdot 25}{2\sqrt{(150-140)^{2}+ 100^{2}}}
(Also, don't use "*" for multiplication in LaTeX. Use "\cdot" instead.)

 

[frosty8688]

I see, my mistake. I must have missed the 2 in the denominator.

 

[frosty8688]

It would be about 21.39 km/hr

 

[eumyang]

That's what I got. You have to be careful with the algebra when solving Calculus problems. (Not easy -- I know.)