Shrinking isosceles triangle The hypotenuse of an isosceles ri...

Question

Shrinking isosceles triangle The hypotenuse of an isosceles right triangle decreases in length at a rate of 4 m/s.

a. At what rate is the area of the triangle changing when the legs are 5 m long?

Accepted Answer

Welcome back, everyone. In this problem, the diagonal of a square-shaped garden is decreasing at a rate of 5 m per second. Determine at the rate at which the air of the garden is changing when the sides are 4 m long. A says it's 10 root 2 square meters per second. B negative 10 root 2 square meters per second. C ne 20 route 2 square meters per second, and D 20 root 2 square meters per second. Now what do we know about the rate at which the era of the garden is changing? Let's start by focusing on that. Recall that the area of a square is equal to the square of its sides. So let's call that Squared where S represents the length of one side. OK. So if A equals Squared, then to find the rate of area change, we can differentiate our function with respect to T. And when we do that, then we get the derivative of A with respect to time, the rate at which the area is changing. To be equal to 2s multiplied by the derivative of S with respect to time, the rate at which the sides are changing, nor do we know the rate at which the sides are changing. Well, not directly, but if we were to draw our square garden here, OK. With the sides S and with our diagonal D, remember, we know that the diagonal of our garden is decreasing at a rate of 5 m per second. In other words, we know that the derivative of D with respect to T equals -5 m per second, OK? Now, since we know that, that means if we can express over one of the length of the sides in terms of D, then we'll be able to use it for the rate at which the diagonal is decreasing. Now, can we do that? Well, we know that we're dealing with a square, OK? And one half of the square will be a right triangle. So we can apply the Pythagorean theorem, OK, to eventually solve for the diagonal, OK. Let me write it above here. We can eventually solve for the diagonal D to be equal to S multiplied by root 2. So since we get that for D, OK, then differentiating the formula for the diagonal D. With respect to T, OK, then we'll get the derivative of D with respect to T to be equal to. Route 2, OK, multiplied by DSDT. Therefore, This tells us that the rate of change of the sides will be equal to, oops, I wrote route S, this should be route 2, my apologies. OK. So route 2. So here our derivative is going to be equal, OK. So. The derivative of D with respect to T multiplied by 1 divided by root 2, and we know that the derivative of D with respect to T is -5 m per second. So the SDT is going to be -5 divided by route 2 m per second. Now that we have that, we can plug it back into our formula for the rate of change for error. As no this implies then that DADT. Is gonna be equal to 2 multiplied by S, which is 4 m or sides are 4 m long, multiplied by the rate of change of the sides, which is -5 divided by route 2 m per second. And now when we go ahead and multiply here, OK. Then we should find that we get the rate of change of area to be -20 multiplied by route 2 square meters per second. Thus, this tells us that the area is decreasing at a rate of 20 2 square meters per second. If we look back at our answer choices, that means is, sorry, D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Shrinking isosceles triangle The hypotenuse of an isosceles right triangle decreases in length at a rate of 4 m/s.a. At what rate is the area of the triangle changing when the legs are 5 m long?

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Shrinking isosceles triangle The hypotenuse of an isosceles right triangle decreases in length at a rate of 4 m/s.
a. At what rate is the area of the triangle changing when the legs are 5 m long?