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# 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 $25 km/h.$ How fast is the distance between the ships changing at $4:00 PM?$(a) What quantities are given in the problem?(b) What is the unknown?(c) Draw a picture of the situation for any time $t.$(d) Write an equation that relates the quantities.(e) Finish solving the problem.

## a) See step for solutionb) $\frac{215}{\sqrt{101}} \approx 21.4 \mathrm{km} / \mathrm{h}$c) see step for solutiond) $2 h(d h / d t)=2(150-x)(-d x / d t)+2 y(d y / d t)$e) $\frac{215}{\sqrt{101}} \approx 21.39 \mathrm{km} / \mathrm{h}$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

So we're giving this problem um and we want to first identify some givens. That's always good because it allows us to know what we're working with. So the first thing that we know is that ship A. This is ship A. Is 150 kilometers west of Shit B. So this will be should be and we know that there's 150 kilometers between them. Then for two we know that ship A. Is sailing 35 kilometers towards east, so this is 35 kilometers per hour east, and then ship B is going um 25 kilometers for our north. And that's our third point. Then for part B, we want to connect those two ideas and we're going to do that through the Pythagorean theorem. So what that looks like is that D squared the distance is equal to 150 the distance minus 35 T. Cause that's the kilometers per hour squared plus 25 t squared. Then we multiply all that out and we simplify then we take the derivative. So what we have is DDT of D squared equals the derivative of are simplified expression 22,500 minus 10,500 t plus 18 50 t squared. We take these derivatives and we simplify and we get the change in distance with respect to time is negative 5 to 50 plus 18 50 t over deep. Um Then with this, since we know DDT we can solve for D. Now choosing our value of T. Going back to the original equation, we get the D squared when T equals four because that's what we're focusing on. When T goes for we get the D squared equals 10,100. So that means that D equals 10 route 101 So that devalue right here is going to be what we used to plug in here to find our change in distance with respect to time. And when we plug in that value, we get the D. D. D. T. Is equal to 21.4 kilometers per hour. Then moving on to part C. What we have is that at noon the distance between the ships is 150 kilometers. So then at four am four hours later, the distance between the two ships is H kilometers. So we see that if this is A and this is B. This right here is 1 50 and then if we have this triangle here with now be being here and a being here. Um the distance here is going to be a church or this is the X. Direction, that's the Y. Direction. So those are our two diagrams and then for part D. We want to analyze those diagrams further. So um we want to find the change in X direction. That's gonna be negative 35 kilometers per hour. That's because this is the um the change in the X direction is going to be a negative movement and the change in the Y direction is gonna be positive, so Dy DT is going to be the 25 kilometers per hour. Um We know that dX cT is negative because four hours later the ship has not reached the point where we started. So what we end up getting is that we want to find the values of X, Y and H. So X. Is going to equal our initial distance minus the distance traveled, so that's 1 50 minus the 35 kilometers per hour times 24 hours. So that's going to give us um that's gonna end up giving us 10 kilometers is the X distance traveled? The wide distance traveled using the same logic is going to be 100 kilometers and then the h distance we found is the distance and that's 10 route, one on one. So then um we have X squared plus y squared equals H squared. We differentiate all this and then what we end up getting is X. DX DT plus Y. Dy DT equals H. D. H. D. T, solving for a D. H. D. T. We end up getting that D HDTV equals one over H times X. DX DT plus Y. DY DT. Okay. And then we have one last part of the problem. That's part E. And what that's gonna look like is um putting in those values that we have now. So we know our value of h we know our value of um DX DT acts all that. So we plug in these values into here. We end up getting that DHD T is going to equal 21 39 kilometers. Okay, our, and that's very close to what we had before.

California Baptist University

#### Topics

Derivatives

Differentiation

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

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