{Use of Tech} Beak length The length of the culmen (the upper ...

Question

{Use of Tech} Beak length The length of the culmen (the upper ridge of a bird’s bill) of a t-week-old Indian spotted owlet is modeled by the function L(t)=11.94 / 1 + 4e^−1.65t, where L is measured in millimeters.

a. Find L′(1) and interpret the meaning of this value.

b. Use a graph of L′(t) to describe how the culmen grows over the first 5 weeks of life.

Accepted Answer

Welcome back, everyone. In this problem, the growth of a certain plant species over time can be modeled by the function G of T equals 18.72 divided by 1 + 5 E to the negative 2.2 T, where G is the height of the plant in centimeters and the T is the time in weeks. We want to use the derivative of GFT to describe how the species grows over the 1st 6 weeks of life by drawing its graph using a graphing utility. Now, if we want to figure out how our plant species grows over time, OK, then we'll need to first figure out the derivative of G of T, put it on a graphing utility, and then interpret the graph. So, let's go ahead and differentiate G of T, OK? So we're gonna differentiate GFT with respect to T. Now notice here that GFTs are quotient, so we can use the quotient rule of differentiation. And here In the notice here that 4G of T recall that the quotient rule tells us that if we have a function Y that's equal to U divided by V, then the derivative of Y will be equal to. Minusuv divided by V2d where U and V are the derivatives of U and V respectively. In this case, our numerator, you will be 18.72, OK? And our denominator, V will be 1 + 5 E to the negative 2.2 T. So when we differentiate those, the derivative of 18.72, a constant is zero, while the derivative of our denominator is going to be -11 E to the negative 2.2 T. OK, we, we differentiate our numerator and bring it down to multiply by our coefficient of E and the derivative of 1 would be zero because it's a constant. Therefore, this tells us. That the derivative of GF T by applying the quotient rule. Actually, let me put that in red here. Is going to be equal to. V multiplied by U. So that's gonna be V multiplied by 0, which would just be 0, minus UV. OK, so that's gonna be -18.72 multiplied by -11 E to the negative 2.2 T. OK. And all of this will be divided by 1 + 5 E to the negative 2.2 T squared. And if we simplify that even further. Then it's going to be positive, OK, multiplying to negative terms. So it's 205.92 E to the negative 2.2 T. Divided by 1 + 5 E to the negative 2.2 T squared. This is the derivative of GFT. This is the function that tells us how the species grows over time. Now that we have this function, the derivative here, we can input it into our graphing utility, in our graphing calculator. I'll leave that for you to do on your own, but you should get a graph that looks something like this one, OK? So this is going to be the derivative of GFT, the graph of the derivative of GFT. Now by looking at the graph what can we tell? We'll notice here that initially, the plant starts growing very quickly, OK? And it reaches its fastest point when T equals 0.73 weeks, OK? Now, at this point here, after T equals 0 points every 3 weeks, the growth is continuing but at a slower pace. Notice that when it gets to week 6, OK, when it starts to get, let me highlight that here. When it starts to get to week 6 and beyond, it gets really close to 0. It starts to approach 0, and that indicates that the plant is nearing its maximum height. So if we were to annotate our graph here, we can tell that here. Initially, it's growing quickly. OK. It's at its max point when T equals 0.73. And then approaching week 6 and beyond. It starts to the growth rate approaches 0, which means that it's nearing its maximum height. That's what the derivative of G tells us about the growth of the plant. Thanks for watching everyone. I hope this video helped.

Key Concepts

Derivative

Graphing Functions

Exponential Functions

Watch next