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The value of a car can be modelled by the function;300V = 18sin0<1<257coswhere, Vis the value f the car in thousands tis the time in years since the car was p...

Question

The value of a car can be modelled by the function;300V = 18sin0<1<257coswhere, Vis the value f the car in thousands tis the time in years since the car was purchased.Express I8sin 0 + 7cos0 in the form R(sin0+a) where R > 0 and 0 < & Give the exact value of R and the value of & in radians, to 2 decimal places_Find the time, in years, that it will take for the car to be worth half of its original purchase price_Find the time taken for the value of the car to reach its minimum

The value of a car can be modelled by the function; 300 V = 18sin 0<1<25 7cos where, Vis the value f the car in thousands tis the time in years since the car was purchased. Express I8sin 0 + 7cos0 in the form R(sin0+a) where R > 0 and 0 < & Give the exact value of R and the value of & in radians, to 2 decimal places_ Find the time, in years, that it will take for the car to be worth half of its original purchase price_ Find the time taken for the value of the car to reach its minimum value. Explain with a reason why this model is not valid for large values of t. (Total for Question 7 is 10 marks)



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A new car worth 24,000 dollars is depreciating in value by 3000 dollars per year. The mathematical model $$y=-3000 x+24,000$$ describes the car's value, $y,$ in dollars, after $x$ years. a. Find the $x$ -intercept. Describe what this means in terms of the car's value. b. Find the $y$ -intercept. Describe what this means in terms of the car's value. c. Use the intercepts to graph the linear equation. Because $x$ and $y$ must be nonnegative (why?), limit your graph to quadrant I and its boundaries. d. Use your graph to estimate the car's value after five years.

Were given a mathematical model. Why equals negative 5000 X plus 45 1000. And we're gonna use this model knowing that why represents the value of a car and X is the number of years and that this car depreciates by 5000 every year. And by finding the X intercept, the Y intercept and graphing the line of this equation, we can find out information about this particular situation or this model and this particular car. So let's start with the X intercept. So the X intercept is the point on my line that crosses the X axis. But it's also the point where why is zero So I can solve my equation with why equaling zero and find X. So let's do that first. So negative 5000 X plus 45,000 equals zero. If I add 5000 X to both sides, I have 5000 X equals 45,000. The next step would be to divide both sides by my 1000 and we are left with X equals nine. So my ex intercept point is 90 We're gonna come back to that in a second. But let's first find Oh, are why intercept? So with our Y intercept, the value of X is zero. So if I take my equation and I put zero in for X, I'm left with why equals 45,000. So my why intercept point is zero. 45,000. So let's take this information and start to plot our line and then talk about what this means. So I'm gonna stay in the first quadrant because I'm talking about a vehicle and I'm talking about time. And so my X axis is in years and I'm not. I can't go back in time to talk about my vehicle. So it's going to stay in the positive and we're not going to go negative and our vowed dollar value of our vehicle either. So my why access is the value of the vehicle dollars and again, I'm gonna stay in the positive. So our years we'll just go up Teoh our X intercept point, which is the nine zero and, well, let's put in parentheses here that this is in thousands. And then we could just go by 10 15 20 2030 35 40 and 45. Save us a little time and labeling our graph here. But knowing that the surface that's 5000, 10,000, 15,000 and so on and so forth. So then let's next plot are ex intercepted or why intercept. So our X intercept is at 90 So put a point here and my why? Intercept is at 0 45,000 So that's right here. And then I'm going Teoh, connect those two points with a line and then we can talk about what this means. So if I look at my graph at the $0.0.9 years by what X intercept means that, um so let me re phrase that so after nine years, the cars value is $0. Okay, and then what is our Y intercept being so at zero years, we haven't even driven the car off the lot yet, right? The value of the car to start is $45,000. So we started at $45,000 before any time passed before we drove it before it started to depreciate. Okay, now let's say we want to know roughly how much is our car gonna be worth after five years? OK, so we can come over to our graph and go to five years and then see about where it's gonna be on our line, and that's roughly $20,000. So we can say after five years, the cars value is about $20,000. And that's how we can use this graphic representation to answer questions about our mathematical model. And that's what's so nice about the representation of a graph, and there you go.

Okay, so before we jump into this problem, let's make sure we understand the context of this problem. So what I'm giving is how much about the value of the vehicle is after T. Years given the original value R. Is the depreciation rate of the value and t. Is how many years after it has depreciated? So Note that the yearly depression rate is 20%. So when I plug in 20 for our no make sure this is converted to decimal divide by 100. So I'm going to plug in 1 -0.2. I'm looking for how many years, how many years And trying to solve her teeth. I don't know what it's original value is but I wanted to depreciate to half of its original value, half is 50% or .5 of times my original value be not. So technically I can divide by the knots both sides and I don't need to have my original value. So I have .5 equals 1 -0.2 to the teeth power Technically 1 -0.2. This is the same thing as 0.8 to the teeth power. And my question becomes, how do I get this T down from the exponent? Well when I had a base of E I could take the Ellen and cancel. But since I don't have a basically here I and I don't have a base in 0.8 I could take the Ellen on both sides. Or I could take the log base 10 on both sides. And by log arrhythmic rules this tea will come down and multiply in front. Such a sense as an exponent. And then to get T by itself divide by Eleanor 0.8 both sides. And let's plug the center calculator in round to the nearest year To the 0.1 of a year. So t. is approximately 3.106, so to the nearest 10th, that zero will round up that one, so it'll take about 3.1 years. For this car did appreciate and lose half its value, which is not much time at all.

Okay, so there's a vehicle that is worth $25,000 and every year it gets depreciated by $2000. What that means is that the overall value is going to be this initial starting amount minus $2000 every single year that you own the car. So this is the formula to find the value of the car at any time, T or T. Is in years. Now. In addition to the value depreciating, the cost is going to rise. So in addition to the value, the cost function says that every single year you own the car, there will be $1500 and miscellaneous repairs and manufacturing and tires. And I don't know, maybe gas. Okay, so it's $1500 every single year. It's own party is also uh in years. So, if you were to graph this, what you would have is on the one hand, this value functions. This is the value function starts at some high amount 25,000, 25,000. And then this negative sign indicates that. Well, this is a linear equation, first of all, because it's in a single variable raised to the first power, and then this negative sign denotes that it's going to have a negative slope. Okay, So it's going to slope downwards at, say 2000. Okay, so I'm going to keep traveling and eventually the car is not going to be worth anything after some amount of time. Okay, so this is the time axis. Okay. Now, in addition to that, we have this cost function I'll do in green, which says that initially the car doesn't cost anything. And then every single year it trends upwards with a positive slope of 1500. That's slightly less in magnitude than the 2000. So, I'm gonna exaggerate this just for the purpose of this problem. So it might look something like that. Yeah. Now this point, right here are those lines intersect. That's a really good um point at which you can sell your car because anything beyond that, anything in this region right here is where the cost exceeds the value. So that means that it costs more to maintain your car and keep it as it is, then you would get by selling it. You pretty much have to keep it and you can never sell it back at this point because you wouldn't get as much money back. So ideally you want to find the time when the value function equals the cost function and then any time less than that time, less than this time right here would be a good time to sell it. So that's one way of doing it. So, if I set these equal to each other, the first thing that I notice is that there's a lot of zeros in here and I don't like zeros. So I'm actually going to divide this entire equation by 1000. Okay, so that means I divide both sides by 1000 and I get rid of all these pesky zero. So I'm left with 25 minus two T equals 1.5 T. And suddenly, in my opinion, that's much better. So I'm gonna add to tea to both sides. Okay? And I get at 25 equals 3.5 t. Now 3.5 can also be written as seven halves. And the reason I'm writing it like that is so we can solve this good old fraction style by multiplying by two. Okay, so right here, that's going to be two times 25 is 50 equals 70 and then divide by seven on both sides. And you get the tea as you go to this in years. Okay, well that's seven and 1/7 years and this 17 translates to about 1.7 months. You just take 12 7 some type that into a calculators, that's about seven years and I don't know 1.7 months. You can convert that in two days if you want, but I'm not gonna do that here. Now another way, uh that the problem mentions is you just take 6% of the starting value. Look at how much time it takes to depreciate to that value. Okay, so what that looks like by the way? Okay, so we have our our 25,000 25 K. And then it essentially depreciates in value. Now I take 6% of that and I'm going to get a horizontal line like this and essentially asking, well at what time does it depreciate? Does it go down, does it go down to, it's only 6% of its value. The first thing I'm gonna do is I'm going to take 25,000 or 25 and I'm going to take 6% of that and look what that is. Okay, so that's 61 hundreds. And of course I'm writing it like this because um this cancels okay and I'm loved with 1.5000. So I want to find out the time when my original function minus um when this function goes okay, I'm sorry the case make it confusing. K means 1000. But as we can as we already discussed we can just divide the whole thing by 1000. So it doesn't really matter. So we have the same kind of equation, but now we're setting it equal to 1.5. Okay, so the value is now a fixed amount. It's 6% of whatever the starting was. Okay so this is a 94% drop. We're looking at how much time elapsed before this happens. So it's another equation that we have to solve. So add to tea to both sides. Subtract 1.5 from both sides. You get that two T equals 23 a half and dividing by two on both sides, they get that T equals um Let's see 11 point 75 Which is 11 plus three quarters years. Okay well clearly 11 is 11. So this 11 years, 3/4 of a year. If you multiply that times 12 to get the months is nine months. So this is nine months. So the final answer for this part is 11 years and nine months.

Today, we'll be using an equation and derivatives to determine the value of an automobile. And that equation for this automobile is V. F. T. Equals 30 times 300.85 to the power of T. Where T. Is time in years and V. Is the value and thousands of dollars. So, first for Part eight, it asks us to evaluate via four. So for that we'll just plugging forward to the equation. So we'll have 30 time 0.85 to the fourth. And using a calculator that is equivalent to about 15.66 And this is in thousands of dollars for part B. It asks us to evaluate or find any equation for the derivative of the F. T. And all right. VF T. Again down below. So we can see it. Uh, Pfft equals 30 times 300.85 to the T. So for this, uh, to find the prime of T any uh, constants that are outside of the part meaning the 30 we'll just stay the same. So 30 will be most the derivative. We multiplied by 30 for this. And now based on uh the formula for the derivative of an exponent like this, we will have natural log of 0.85 times 0.85 to the T. And that will be older if if you can simplify it if you like. But essentially that is what it is. And then for part C. It asked us to find the promo for. So we'll just plug in uh four for v prime ft. Still have 30 times Ln of 0.85 Yeah, times 0.85 to the four. And using a calculator that is equivalent to about negative 2.545 Now, based on these two numbers to be prime before and view of four and both of these are thousands of dollars by the way. Um Based on the numbers. Uh this negative right here indicates that the value goes down over time. Uh It goes down by 2.5 ish over four years um in thousands of dollars. And it starts off as 15.66 thousands of dollars. So for part D. Were as uh if these numbers support or oppose the fact that you should keep their car as long as possible, um even though it depreciates, it doesn't appreciate that much. So you should keep it as long as possible. So you don't have to keep buying new cars over and over again.


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