Once Kate’s kite reaches a height of 50 ft (above her hands), ...

Question

Once Kate’s kite reaches a height of 50 ft (above her hands), it rises no higher but drifts due east in a wind blowing 5 ft/s. How fast is the string running through Kate’s hands at the moment when she has released 120 ft of string?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says a weather balloon is tethered to the ground and reaches a steady altitude of 80 ft. It drifts horizontally due to a wind blowing at 6 ft per second. How quickly is the tether being let out when 150 ft of tether has been released? We're given 4 possible choices as our answers. Your choice A, the tether is being let out at -6.08 ft per second. For choice B, the tether is being let out at 6.08 ft per second. For choice C, the tether is being let out at minus 5.08 ft per second, and for choice D, the tether is being let out at 5.08 ft per second. So the question is asking us how quickly the tether is being let out. 150 ft of tether has been released. Now we're also told that the balloon is reaching a constant altitude of 80 ft, but there is a horizontal wind that is blowing at a rate of 5 ft per second. So this means that our tether is actually the hypotenuse of a right triangle. So we're going to draw that right triangle just to visualize what we're given in the problem a little bit better. So we're going to draw a horizontal line indicating the ground, then we'll have a dashed vertical line indicating our altitude, and we'll label that distance as H. The wind is going to be blown horizontally, so that's going to displace our balloon in the horizontal direction. So we'll draw a dashed horizontal line to the right, and we'll label that distance as X, and then the pots of that right triangle is going to be the length of our tether, which we'll label as S. So, based on the information that we're given in the problem, we have H, it's going to be equal to 80 ft. Since 80 ft is the altitude given in the problem. We weren't given any information about X directly in the problem. We were told that the wind is blowing horizontally, at a rate of 6 ft per second. So that means our balloon is going to be moving at that rate. In the horizontal direction. So that means that the derivative of X with respect to T. It's going to be equal to 6 ft per second. And then finally, we want to know um the rate at which the tether is being let out when 150 ft of tether has been released. That means that S is going to be equal to 150 ft. So, we're ultimately looking for the derivative of S with respect to T. And so, to begin to find that, we're gonna start off with Pythagorean's theorem. So we call for Pythagorean's theorem that we're going to have S2 is equal to X2 + H2. And what we're going to do is take a derivative with respect to T of both sides of this equation. So on the left hand side, I'll have the derivative with respect to T of Squad. I'm I take that derivative, I'm gonna get 2 S, multiplied by the derivative of S with respect to T. And this is going to be equal to, on the right hand side. I'll take the derivative with respect to T of X2, and that'll give me 2 X, multiplied by the derivative of X with respect to T. Then we'll have plus the derivative with respect to T of H2d. However, H here is constant since our altitude is remaining steady. And so here I'm taking a derivative of a constant, and anytime I take a derivative of a constant, I will get 0. So we'll have plus 0. We can simplify this expression by dividing both sides of our equation by 2, and it cancels out a factor of 2 on each side, and then we'll be left with S, multiplied by the derivative of S with respect to T. is equal to x multiplied by the derivative of X with respect to T. Now, we need to solve this for the derivative of S with respect to T, so we'll divide both sides of this equation by S, and we'll be left with the derivative of S with respect to T. is equal to the quantity of X divided by S in quantity multiplied by the derivative of X with respect to T. Now we have values for S as well as the derivative of X with respect to T, but we do not have a value for X. However, we can again use Pythagorean's theorem to determine that value. So, solving our equation for um Of Pythagon's theorem for X2, we'll get X2 is equal to S2 minus H2. Now we just need to take the square root of both sides, so we'll have X is equal to the square root of the quantity of S2 minus H2, and now we just need to substitute in values for S and H, so this is going to be equal to the square root. Of the quantity of 150 ft. Squared Minus 80 ft Squared And evaluating those values inside the square root sign, this is going to be equal to the square root. Of 16,100 ft squared. And then using our calculator to find the value of this square root, this is going to be approximately equal to 127 ft. So now we just need to substitute in our values for X, S, and the derivative of X with respect to X into our expression, or the derivative of S with with respect to T. So, doing so this is going to be equal to. The quantity of X, which is 127 ft. That is divided by S, which is 150 ft. In quantity Multiplied by the derivative of X with respect to T, which is 6 ft. Per second So evaluating this expression, this is going to be equal to 5.08. Beat per second. So that is the answer that we were actually looking for. And we compare that to the answer choices, we see that the correct answer choice corresponds to answer choice D. So I hope that this explanation has been useful, and I'll see you in the next video.

Once Kate’s kite reaches a height of 50 ft (above her hands), it rises no higher but drifts due east in a wind blowing 5 ft/s. How fast is the string running through Kate’s hands at the moment when she has released 120 ft of string?

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