{Use of Tech} Power and energy Power and energy are often used...

Question

{Use of Tech} Power and energy Power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up. It is measured in units of joules or Calories, where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶J, or 240 Cal. On the other hand, power is the rate at which energy is used, which is measured in watts, where 1 W=1 J/s. Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ W). If energy is used at a rate of 1 kW for one hour, the total amount of energy used is 1 kilowatt-hour (1 kWh=3.6×10⁶ J) Suppose the cumulative energy used in a large building over a 24-hr period is given by E(t)=100t+4t²− t³/9 kWh where t=0 corresponds to midnight.
a. Graph the energy function.

a. Graph the energy function.

Accepted Answer

Hi everyone, let's take a look at this practice problem dealing with derivatives. This problem says, energy and power are terms frequently used interchangeably, although they are slightly different. Energy is defined as the ability to provide heat or to do work, while power is the rate at which energy is used. The total energy consumed of a residential building over 24 hours is given by the following function. EFT is equal to 200T plus 4.5 T2 minus T cube divided by 8, where T equal to 0 corresponds to midnight. We need to provide a graph of the energy function. Now, to draw this graph by hand, we're first going to need to determine where the extrema are. And so, we call to determine where the extrema are located, we need to take the derivative of our function and set it equal to 0, and then solve for T. So we're gonna be looking at the derivative of E with respect to T, we'll be looking for E of T. And here, when we take the derivative, we're actually gonna be using our power rule for derivatives. So, we call for the power rule that we have the derivative with respect to X of the quantity of X raised to the N, and that is equal to N, multiplied by X rated to the N minus 1. So, applying this rule to the function that we have for E, we will have the derivative of 200T with respect to T, which when we use our power rule, it's just going to be 200. Then we'll have plus, we'll have the next term is 4.5 T squared, and we take that derivative, we multiply the 4.5 by 2, which is the exponent, which will give me 9. And then we are multiplying 9 by T. Since we'll have 2 minus 1, which is gonna be 1 in our exponent, but T to the 1 is just T. Then finally, for our last term, we'll have minus the derivative of T cube divided by 8, and that is going to be 3 T 2 divided by 8, again applying our power rule. What we need to do is set this equal to 0. When I set this equal to 0, I noticed that I have a quadratic function, and it's already in the quadratic form. So we can actually use our quadratic formula to determine the values for T. And so we call for the quadratic formula we have T is equal to. 9, plus or minus the square root. Of the quantity of 92 minus 4 multiplied by 200 multiplied by Minus 3 divided by 8. And then that whole quantity is divided by the quantity of 2, multiplying the quantity of minus 3 divided by 8. So now we need to simplify this expression. So doing so, we'll have T is equal to. -9 Plus or minus the square root of the quantity, 92 is 81, and when evaluate the minus 4 multiplied by 200, multiplied by -3 divided by 8, I get plus. 300 Then that quantity is all divided by the quantity of minus 3 divided by 4. So we can again simplify this a little bit um more by multiplying the numerator and denominator by -4. So doing so, we'll get T is equal to 4, multiplying the quantity of 9, plus or minus square root of 381 in quantity, divided by. 3. So this gives us our two solutions, and so for T1, we'll take the positive route. So T1 is going to be equal to. 4 multiplied by quantity of 4 of 9 plus square root of 381. In quantity, divided by 3. And this is approximately equal to. 38.03. For T2, which is going to be the negative route that we we're going to take for the square root sign. So T2 is going to be equal to 4. Multiplied by the quantity. Of 9 minus. The square root of 381. In quantity, divided by 3, and this is approximately equal to minus 14.03. Now, both of these values for T fall outside of the domain for a function, which was from 0 to 24. So, I'm not gonna be plotting any of these um extremo points on our graph. So next we need to move to finding any inflection points. And so, we call for inflection points, we need to take the second derivative of E with respect to T and set it equal to 0 and then solve for T, and that's gonna be the locations for my inflection points. So, taking another derivative of E, so I'll have Eme of T. Here we just need to take another derivative of our first derivative E. And when I do so, E of T is going to be equal to, when I take a derivative of 200, it's a constant, so it's derivative of 0. The next term is 90. I'm gonna take a derivative of 90, I get back just 9. And then I will have minus. The derivative of 3 T 2 divided by 8. When I take that derivative using our power rule, we'll get 3 T divided by 4. And to find the inflection point, I need to set the second derivative equal to 0, and then solve for T. So solve for T, we need to just subtract 9 from both sides of our equation. And we'll get minus 3T. Divided by 4 is equal to minus 9. And then I just need to multiply both sides of our equation here by 4. And then divide both sides by -3. And when I do so, On the left hand side, I'll just have T, and that's gonna be equal to, when I evaluate the right-hand side, I get 12. So, our inflection point is located at T equal to 12, and so we need to find the value for E at T equal to 12. So, we'll have E of 12. is equal to 200 multiplied by 12. Plus 4.5, multiplied by 12 squared. Minus 12 cubed. Divided by 8 And what if I with this expression I get. 2,832. So, we will use that value when we actually make our graph. Now, before we can actually make a graph, we need to determine the values at the end points of our domain at T equal to 0 and T equal to 24. So we'll start off by evaluating E at T equal to 0, so we'll have E of 0. This could be equal to 200 multiplied by 0, plus 4.5 multiplied by 0 squared. Minus 0 cubed Divided by 8, and when I evaluate that expression, it's going to be equal to 0. And the next value that we need to evaluate is when T is equal to 24, so we'll have E of 24. And that is equal to 200, multiplied by 24. Plus 4.5. Multiplied by 24 squared, minus. 24 cute. Divided by 8 And so when we evaluate this expression, we get 5,664 So now we have our two endpoints as well as our inflection point. We can now create our graph. So we're gonna plot our graph on A grid? And so we have our X and Y axis shown on the screen. And so we're going to draw label 3 points. The first point we're going to draw is actually at 00. The next point is our inflection point. Which was at 12. Equal 12 and then I need to go up. Until I reach 2,832. So that's gonna be just below our 3000 mark. At T equal to 12. And then our final point is going to be at T equal to 24, and a value of 5,664. So now we've placed all of those points, we need to connect them, ensuring that we actually have an inflection point at T equal to 12. So doing so, We get something similar to This graph here. Where we have an inflection point at T equal to 12. And we're going through the points of 00 and 24 and 500. Oh sorry, 5,664. So this is the graph that the question was asking for, so this is the answer to this question. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Energy and Power

Graphing Functions

Cubic Functions

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