Hauling in a dinghy A dinghy is pulled toward...

Question

Hauling in a dinghy A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is hauled in at the rate of 2 ft/sec.

b. At what rate is the angle θ changing at this instant (see the figure)?

A dinghy being pulled toward a dock by a rope, with a ring on the dock 6 ft above the bow, illustrating related rates.

Accepted Answer

Welcome back, everyone. A ladder 20 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 3 ft per second, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 12 ft from the wall? A says it's 16/3 of a radiance per second, B 31 16th radiance per second, C 3 16th radiance per second, and D, 1/16 radiance per second. Now, first of all, what are we trying to figure out? Well, we want to find how fast the angle between the ladder and the ground is changing given the condition that the bottom of the ladder is 12 ft from the wall. In other words, we want to find a change in theater or angle with respect to time. The theta T. So how can we do that? Well, probably a diagram will help us best here. So let's go ahead and try to draw a scenario to figure out what it's saying, OK. So let's say that this is ROI. OK, and this is our ladder, and we know that the bottom of the ladder is 12 ft from the wall. So let's say that X equals 12 ft. We want to know how our angle is changing, so let's call the angle that the ladder makes with the ground theta, OK. And we can call the length of our vertical wall wide. How can we figure out uh the the theta with respect to time? Well, we'll need to express theater, uh, in terms of the information that we have. And we could use what we know about trigonometric ratios to help us. Now, remember, OK, uh, so far we know that the length of the ladder is 20 ft, so we should actually put that here. And if we recall. That our angle theater is adjacent to the side X. So based on what we know about trigonometry, recall that the cosine of an angle is equal to the adjacent side divided by the opposite side, um, the hypotenuse. It's the ratio of the adjacent side to the hypotenuse. So if we apply that here, then this means the cosine of theta equals the adjacent side X divided by the length of the hyperten use 20. No, remember, we need our change in theater with respect to time, so we can differentiate both sides with respect to time T. In other words, we can differentiate the cosine of theater with respect to time. And a 20th of X with respect to time. OK? So let me write that properly here. OK, so that's just oops, my apologies. respect the time of. X divided by 20. Now the derivative of the cosine is the sign, so this is gonna be a negative sign theater multiplied by D theta divided by DT. And for our other side, uh, the derivative with respect to the T of X divided by 20 is going to be 1/20 multiplied by the XDT, the derivative of X with respect to time, OK? No, that means then in order for us to figure out the theta DT. We'll need the sign of theta and we'll need the XDT. But remember early in our problem, we were told that the bottom of the ladder is sliding away from the wall at a rate of 3 ft per second. So we already know then that. DXDT. is equal to 3 ft per second. So now all we need to do is to figure out the value of sine theta. So how can we do that? Well, again, if we go back to our diagram here, recall. That the sign of an angle, the sign of theater will be the ratio of the opposite side to the hypotenuse. So in this case it's going to be equal toy divided by 20. We don't know what Y is, but if we know the length of our hypotenuse and the length of our base, then we can figure out the height using the Pythagorean theorem because by the Pythagorean theorem. By the Pythagorean theorem. OK. We know that 20 squad will be equal to 12 2 plus Y2. OK. So, Y2 will be equal to 202 minus 122. That's gonna be 400 minus 144. And if Y2 equals 256, then Y will be equal to the root of 256, which equals 16. So the height or vertical height is 16 ft. So now, since that's the case, if we put that back into a range of a sign. And that means the sign of our angle theta will be equal to 16 ft, divided by or hypotenous 20 ft or 45. Now that we have a value for sin theta, we can substitute it into our problem, OK? So, given the XDT equals 3 ft per second and sine theta equals 4/5, OK. Then, That means our problem is gonna be negative 4/5 multiplied by D theta DT. Equals 1/20 multiplied by 3 ft per second, OK? And if we sha the theta DT by dividing by negative 4/5, then we get it to be equal to -320 multiplied by 5/4, OK, which if we do some transposition here, 5 goes into 51 time, 5 into 24 times, -3 multiplied by 1 is -3, and 4 multiplied by 4 is 16. So our angle is changing at a rate of -316 radiance per second. In other words, the negative sign tells us that the angle is decreasing as the ladder slides away. If we look back at our answer choices. Then the C is the correct answer because we're not interested in if it is decreasing or increasing. We just need to know how fast our angle is changing, OK? Therefore Let's make a statement here. The anger between the ladder and the ground. Is decreasing At a rate Of 3/160 radiance per second. C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Hauling in a dinghy A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is hauled in at the rate of 2 ft/sec.

b. At what rate is the angle θ changing at this instant (see the figure)?

Key Concepts

Related Rates

Trigonometric Relationships

Implicit Differentiation

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