You are watching: What happens to the speed of the boat as it gets closer to the dock?

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take into consideration this illustration of the problem listed below. Now and here we want to discover the speed of the watercraft once 13ft of rope is out. And you want to know what happens to the speed of the boat as it gets closer to the dog, offered that the winch pulls in rope at A rate of 4 ft/s. So initially we wanted to discover our variables, let's say L. Is the length of the rope. And let's say accessibility the distance in between the dock and the watercraft have a pythagorean theory. And we have the relationship between L. N. X. That somewhere else. That's equal to X squared plus 12 squared. Or that's L squared equals X squared plus 244. Now, by implicit differentiation we have actually twice L. A. Times D. L over D. T. This is equal to twice X times dx over Bt. And if L is equal to 13 and also dl over details four ft per second. We have 2 times 13 times for This equal to two times X times dx over D. T. Simplifying that we have 104. This is equal to X times dx over D. T. Now we deserve to uncover the worth of X. Using the relationship that we have for L. Index. So if our l. s. 13, then from right here we have 13 squared equal to x square plus Under 44 you have X squared equal to 13 squared minus 144. That's 25 which suggests that X equates to five. So then from here we have actually 104 this is equal to two times 5 times dx over D. T. Which suggests that the X over DT equals 10.4 ft/s. Now, to identify the rate of readjust of the rate of the boat, we desire to get the 2nd derivative of X through respect to T. So we will certainly usage the equation twice L. Time's D. L over D. T equates to 2 X times dx over D. T. You will certainly simplify this initially. From tright here. We obtain L. Times D. L over D. T. That's equal to X times dx over D. T. This is after splitting both sides of this equation by 2. And then from below, we identify this aacquire through respect to T. We have ah we need to perform product dominion both sides. So we have L times derivative of dl over DT that's D squared L over DT squared plus the L over D. T times L derivative of that will certainly be D. L over D. T. This is equal to aget, product dominance that's X times D squared X over DT squared plus we have actually dx over D. T. This times derivative of X. With respect to T. Which is dx over D. T. Now, given that the rate of adjust of the size of the rope continues to be continuous than the rate of that Will be zero. So this is zero. And given that you currently have actually the value for D. L over D T. Which is 4 ft per second. You have 4 times for that's equal to X. Here is 13 times D squared X over DT squared plus DX over D. T. already uncovered this to be 10.4. So 10.4 squared, addressing for destroyed X over DT squared we have D squared X over D. T square. This is equal to 16 -10.4 squared.

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That's separated by 13 and this is the worth equal to -7.0 89 feet per second squared. And because this is negative, the price of the speed yes, decreasing at it as it gets closer to the dark, it provides sense because the boat has to slow-moving dvery own as it gets closer to the dog.