solve-the-given-problems-it-is-estimated-that-a-newly-discovered-oil-field-will-produce-oil-at-the-3

Question

Solve the given problems.
It is estimated that a newly discovered oil field will produce oil at the rate of $\frac{d R}{d t}=\frac{\left(400 t^{2}\right)^{2}}{t^{3}+20}+10$ thousand barrels per year. How much oil can be expected from the field in the next ten years?

Stanley Enemuo · Accepted Answer

so we have the definite integral. Okay Toms what the negative 1/2. See why divided by a Times Radical two. The so now we have to evaluate the integral where we could use the power form of for integration. So you the integral of U. And see you. That&#x27;s equal to you. Times M-plus one Divided by M-plus one plus C. Where an it&#x27;s not equal to negative one to the constancy of integration. When N. is equal to negative one hand, they use equal to Y. So D. You this equal to DY so we have to limit to a Divided by eight times radical G. Times one Homs Wadi to the one hand. So that&#x27;s equal to two a Divided by eight times radical G. Medical to G. So that&#x27;s radical to a developed by A Times Radical G. So eight to the one half minus 8 to the 1/2.

Helena Redgate · Answer

were given an initial rate of consumption and then a growth rate for the change in that consumption. So we&#x27;re gonna note that our exponential growth function will be modeling the rate of change, and we may be using intervals. In this problem were asked to find the amount of oil consumed in 2010 where our time is zero and then the general function for oil consumed between that T zero value and any future time. And finally, the time when the amount of oil consumed is 10 million barrels. We&#x27;ll take the 1,000,000 off our numbers here and be conscious of adding it back as part of our units. Okay, so part A is where we will start, are right function Looks like this are of t is equal to are not the initial rate times e to the k t. And we need to find this k value. So we know that our of one from our growth rate that we were given that will be 1.15 for 1.5% times are initial value is 1.2. So that&#x27;s gonna and that&#x27;s also equal to when we plug in our values to this equation here. So 1.2 e to the K. And to use one. So we&#x27;ll cancel our 1.2 and to solve for K. We&#x27;ll get that. The natural log 1.15 is equal to okay. And in this case, it will be a good idea to approximately that easy a calculator. And that is 0.14 none. So our equation looks like this are of T is equal to 1.2 e to the t times the natural log of 1.15 And we have to other ways of writing this because this tease outside the natural log, we can send that to the exponents in the natural log can cancel. So we could also write it as 1.2 times 1.15 to the T. And since we approximated are natural log, we can also write 1.2 e to the 0.1492 All right now, what we want to take is the integral over the first year. So that is 0 to 1. Because our t zero is your years and r T one is one year of the rate function times or with DT as a closing for integral. Always important to remember that. And this is equal to the integral from 01 of 1.2 e to the 0.149 t d t. And we can solve this using what we know about E with an exponents in integral, this will be we put the constant 1.2, and then we will get that over 0.149 times E to the 0.149 t evaluated 01 That&#x27;ll be one point 2/0 1.0.1 or nine times E to 0.149 minus one because anything to their powers. One. And this is approximately 1.21 million. That&#x27;s where units are important. It&#x27;s the integral because we&#x27;re modeling the rate of change with this function. And so if we want to know the actual amount used, that&#x27;s why we take the integral or the area under the curve is another good way to think about it. Okay, that&#x27;s part a move on to part B to find the consumption will call a new function See of tea. And we have a serum three in section 6.1. This the room will give us a future value. And so that looks like this See, of T is equal to see of zero plus the integral from zero to t of r of s evaluated over us and as here is just a dummy variable. So we&#x27;re not using t in two ways at one time. And we can do this because our is the change in si because we are calling CR consumption. Right? Okay, so our consumption zero is going to be zero trust this integral zero to t of 1.2 e to the 0.149 s d s. And we can evaluate this much like above in part a. So this will be one point to the constant over evaluating e with exponents 0.1 for nine. This is times E to the 0.149 s, evaluated from zero to t and we will use our calculator to evaluate one point 2/0 1.0.149 and we&#x27;ll get 80.59 and then this in here we evaluate a TV minus evaluating at zero. That&#x27;s E to the 0.1 for nine T minus one. Anything through the fear of power is one. Okay, so that&#x27;s part B. That will give you what the question asks for the function. We can move on to part C. We want to determine when the consumption is 10 million. So we want see of t this you function. We defined to be 10 and so we just have it right here. So 10 equals 80.59 e to the 0.149 t minus one, and we&#x27;re solving for this tea here. So first, we&#x27;re gonna divide by this constant, and then we&#x27;re gonna add this constant here, and then we&#x27;re gonna take the natural log. And so what we&#x27;ll end up with is 0.1 for nine T equals the natural log of 10 over 80.59 plus one, and then we&#x27;re clearly going to divide by 0.1 for nine on both sides. We&#x27;ll end up with T equals 7.9 years

Carson Merrill · Answer

So for this problem, we are going to be looking at oil consumption. Uh, part is asking us how much oil was consumed over the course of the year. 2018. Um so to figure this out, we first need to get a a graph of the function. Um, so in order to use, we need our K value and the K value is going to be the natural log of one plus the rate. So that would be the natural log of 1.15 It&#x27;s other K value. Um, we make this function, it&#x27;ll be wife Kia to function Time is equal to the initial value, which is 1.2 million barrels. So 1.2 uh, E to K, which is 0.1486 t. So there is our, um, function. Um, and we want to now determine how much oil is consumed over the course of the first year. Eso To do this, we are going to using integration. To find the area under the curve, eso will use the integral tool from zero toe one, and it is going to be this function right here with respect the time. And what that tells us is that over the course, the first year one approximately 1.21 million barrels of oil were consumed in that first in that one year of 2018 and then we&#x27;re going to be finding a function. Uh, that tells us the amount of oil consumed between zero and a future time. Uh, this would essentially be, uh, this function right here. Um, as long as we can integrate with respect to tea after take the anti derivative of this function right here, what you&#x27;ll find is a general function for the total amount of oil over the course of any amount of years. Um, so then we&#x27;re being asked after how many years of 2018 will the amount of oil consumed reach 10 million barrels? Um, and if you have the If you&#x27;ve computed the anti derivative, um, you tend just plug this into your function. Um, if you don&#x27;t compute the anti derivative, we can plug in some values and we&#x27;ll see that at eight years it reaches 10 Merrill 10 million barrels. About it. Seven years. It doesn&#x27;t so 88 years. Um, within those eight years, it does reach 10 million barrels, but not within seven. So between by the seventh and the eighth year and within the problem, that would be after, uh, S o b. Seven years after 2018. So after in the year 2025 um, it will reach 10 million barrels and then by 2026 it will always have over 10 million barrels of oil consumed.

Matt Just · Answer

all right. Uh, the rate of consumption of oil and billions of barrels by, um See, that&#x27;s cos given by okay, rate of consumption. So unless you see for consumption. So if we have the rate of consumption, we have C prime of T is 1.2 year 20 point 04 teeth where t equals zero corresponds to the year 2010. And we want to find the total amount of oil used by the company from 2010 to hear t So we went see of T minus C, that&#x27;s zero, which is just a change from zero to t of dysfunction 1.2 e to the 0.4 t t. T. Okay, and we just need an anti derivative 1.2 is a constant anti derivative of this is just itself, but we need to divide by 0.4 to account for when you take the derivative, we&#x27;ll get a factor zero point here before and then this function that we get, we just evaluate from zero to T, which is 1.2 over 0.24 mhm he to the 0.4 capital T minus E to the zero, which is one. So there&#x27;s the function and then it asks at this rate, how much will be used in five years? So we just plug in two equals five into this formula. So when t equals five years, we get that. The net change is approximately 6.64 billion Paris.

Aishwarya Krishnakumar · Answer

problem 71. We are given this oil consumption word problem. So for part A, it asks us to write c prime of tea for the oil company Leading T equals zero. So for part A, we would just rewrite this except Artie would be zero would be c prime of tea because so the cave value is 1.2 I was given in the problem. That would be 1.2 e to the and the R value is 0.4 and t you would just have here and for part B. It asks us to set up a definite inter grow for the amount of oil that the company will sell in the next 10 years. So that means our inner girl bounds would be 0 to 10. This is going up to 10 years, and we would just have the sea of tea inside, so 1.2 into the point before T d. T and part bees just asking us to set it up, so this would be a sufficient answer. So part see asks us to evaluate this in a girl. So to do that, we will just we don&#x27;t we don&#x27;t need any u substitution. We just look at this and the integral off e to the point. 04 t would just be need to the point. 0 40 over point. 04 So, me to the point, 04 t over 0.4 And don&#x27;t forget to bring down the 1.2 and this would be evaluated from 0 to 10 which really does make it easier, because if you plug in 10 you get a value. But if you plug in zero, the top would cancel out. So just be one over 0.4 So we have 1.2 times e to the 0.4 times 10 over to your point here for and then if you plug in the zero, you would get one over I know for and this gives us about 14.7. This is in billions and for party. It asks us to do everything the same. Except now it goes up to um so now it has 20 billion barrels of oil. So it has. We have to find the tea value. So we have to find this t value. So 1.2 you toothy you know for team, and it says it adds up to 20 billion. So we can we know what we got for the integral, which was this value over here. So if he&#x27;s if you, um if you simplify this, you would have gotten 30 times me to the point for T minus one. So now you can just go ahead and solve for tea. So if you saw for tea we get t equals approximately 15 years and finally for part E. We&#x27;re asked to do the same thing that we did for party. Except we have to assume that our equals 0.2 instead of 0.4 So we do the same thing. Except this would be 0.0 two. So if you do the exact same thing letting t letting Ari quotes point or two. You can again Saul for tea, and you get a new value for Team, which is approximately, or tea

Jillian Rae Villa · Answer

the question as to solve are given the equation a equals p Times EADS is a power of RT than to find the rate of which $1000 deposited with double in three years. So given that a equals p times e r T, we can isolate our by first dividing a by P. Then we can take the natural log of both sides, and we take the natural log because the natural log off e to any power or to any exponents equals that exports. So now the equation becomes Ellen off a over p equals R T. Then we divide both sides by t and we get a overpay divided by T equals R. Now that we have the equation, we want to find the rate at which $1000 deposited would would double in three years. So that means at the end will have $2000 because 1000 times two is $2000 in this equation, T ever. But it represents the amount of years or three years A is our future value, or $2000 P is a principal value or deposited value, which is $1000 so plugging in the values we have Ln of 2000 divided by 1000 all over three years, or Ln of two divided by three, and we would get 0.231 or 23.1 percent.

Problem

Solve the given problems. The work $W$ (in $\math…

Problem 48 Easy Difficulty

Solve the given problems.
It is estimated that a newly discovered oil field will produce oil at the rate of $\frac{d R}{d t}=\frac{\left(400 t^{2}\right)^{2}}{t^{3}+20}+10$ thousand barrels per year. How much oil can be expected from the field in the next ten years?

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Precalculus Review - Intro

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Basic Technical Mathematics with Calculus

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Catherine R.

Caleb E.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus