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For the following problems please show all vour KurCEecem full crexlic _hr)Find R(r) Simplily pible. (Hint : Dot tQuotient Rule)Find & ((r + 1)(2} Tr] 10))Iutr ...

Question

For the following problems please show all vour KurCEecem full crexlic _hr)Find R(r) Simplily pible. (Hint : Dot tQuotient Rule)Find & ((r + 1)(2} Tr] 10))Iutr Ltv = Final v" 7tar+3f(r) = Vsin(v7) Find f'(T)Find W""(=)W(t)

For the following problems please show all vour KurC Eecem full crexlic _ hr) Find R(r) Simplily pible. (Hint : Dot t Quotient Rule) Find & ((r + 1)(2} Tr] 10)) Iutr Ltv = Final v" 7tar+3 f(r) = Vsin(v7) Find f'(T) Find W""(=) W(t)



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Find an exact solution to each problem. If the solution is irrational, then find an approximate solution also. Height of a Sky Diver A sky diver steps out of an airplane at 5000 ft. Use the formula $S=-16 t^{2}+v_{0} t+s_{0}$ to find how long it takes the sky diver to reach $4000 \mathrm{ft}$.

All right. And this problem we wish to find DZ DT the director of the with respect to T. Using to enroll one where Z is defined parametric lee as L n X, Y where X equal to the fifth and Y equals root T plus one. This question is challenge your understanding of both Highbury a differentiation to solve, we rely on chain rules are functions of several variables. Were chain rule one states that D C T and G d X dx DT plus DZ Dy Dy DT to solve the therefore must find the four derivatives given in the formula for the cgt. So differentiating and the first we obtain D C D x equals Y or X. Y equals one over X. DZ dy analogous lee is excellent. X Y equals one over Y. Differentiating X and Y. Now expected gives the X c t equals five to the fourth. Dy DT equals 1/2 plus one. Plugging in D c d t. Gives the formula on the left. Where by plugging in X and Y, we obtain solution. DZ DT equals five over T plus 1/2 times two plus one as a shaded on the right.

In the problem we have been given W equal X. Y. Plus. Why said plus X. In jail this is the extent now X is equal to the square, Why is equal to 1 -1 sq And there is equal to one -T. So The except on duty is equal to two. T. Do you buy upon DT is equal to -2 T. And dessert upon duty is able to -1. So do you w upon DT equal? Why do you accept on duty plus X? They were up on duty plus injured, develop on duty plus away dessert upon duty plus jerry dx upon duty let X. These are upon DT. Further this equal one minus the square into two. T. Blessed describe into minus duty Plus 1 -9. Into minus duty plus one minus D squared into minus one Plus 1 -9 into two T blessing. Do you square into -1? Further after simplification we obtained this size -4 due to the power three Plus two T -1, so D w upon DT equal minus 40 to the par three plus two t minus one, and this is the answer.

The rate differential of a very respected T. In standard pollution index per hour of a pollutant put into the air by a smokestack is given by differential wave respected t equal 150 over one plus 0.25 Times T -4 sq and all that plus 25 40 years time. In our after six a.m. We're going to estimate the total amount of pollutants put into the air between six am and noon is interrupts of the rule with six some infamous. So if this is the rate of pollute and put into the air then the Total amount of polluting put into the air between six and 9 will be the integral from 0 to 6 of this expression here and zero from 0 to 6. Because at zero we have zero hours after six a.m. Because we start just at six a.m. And at noon we have six hours after six a.m. So let's say that first. So you want two estimate Angel from 0 to 6 of 150 over one plus zero point 25 times T -4 sq last 25 by the episode. The rule with six sub intervals. Mhm. So we define the function f of t equal the integration here is 150 over one plus 0.25. T minus four square last 25. That's functional. 15 be defined On the interval 06. So oh then we calculate age which is the step size. That is a common distance between any two consecutive notes and ages defined as the length of the interval of integration 6 0 over The numbers of Interval six. That is one. So step sizes one with this step size we can say the notes are it's not equal zero x 1 equal one X two equal to X three girls 3 x four equal 4 X five equal five and X six equals six. And now we can say that the integral from zero 2 6 of the function. Fft they find here above Is approximately equal to T6. There is a trap cider rule with six of intervals which is fine as age half times the image of the first note. It's not and here maybe I'm going to change something and is that we are using the variable T. So maybe it's better he is better T instead of X. Doesn't change anything and very clear but he turned tough Consistency of meditation. So here is the zero T. 1 T two T. Three 84 T five and T. six. Okay So it's f. at the zero plus two F. At T wan plus two. F at T. Two Was to f. 33. List to f. Plus two f. T five Plus F 86. Just put the six better. He also saw statistics here. Okay is it let's say we arrange this you better left. Okay and this is equal to ages one so one half times F zero. Mhm Let's do F at one is to effort to was to effort three blessed to if at four was to effort five plus 76. This us one half times. Now we relate the function of this virus. Ft zero Yeah 150 over one plus 0.25. four square first course 16. Now we have 16 Time 0.25 gain and say F to your room here's five. Okay This is 55 for those two times. If at one yes then it's not equal but approximately equal because I'm here approximations to the number Of the image 71 point 1538. Okay. 461 538 4616. There is only this one here plus Two times f at two is 100 200 here. Yeah we can put that lows 200 last two times if a three 100 45 times 290. So with that also here us 290 plus. So we have a fat 1234. Yes effort for is 175 times two. 115 plus At five we get Okay. Uh huh. 45 that is again 290 plus at six Andrea 100. So except F one just an approximation with all these things. It's all the numbers. All other numbers are interesting numbers so could have been an interview but Because of F one we have an approximation so that's approximately equal to. We do all these calculations. Now let's see the number into your number. Maybe first Yes one half Yeah. Times. And now some of the interior numbers here is 1285 plus. and now two times this number here let's say is equal to 1 42 were you point 3076. Okay. 33% 69,230 769 mm three. Now we made some here. That is um 142. 142 points. Sorry we're no it's not. That is yes 40 1427 1427 point three Series 7 six. Mhm. Mhm. And two three 076 uh nine to and now this divided by two is equal to it's about 700 13 point 65 38 4615. Okay. 384. Okay 62. Mhm. And this this should be in a standard pollution index. So uh we can say that The intervals from six For its from 0 to 6 of F of t differential T. So approximately equal to 307 100 13 0.6 five Let's say 4 to stay with six figures are or digits and three decim. And then this implied that implies that the total amount of pollution put into the air total amount of for written the victors. How for you to food into the air Between six am and noon is about 713.6 54 pollution index. Standard pollution index. Yeah. Mm. And that is the answer. We have calculated numerically by using traps out of the wall with six of intervals and the internal was integral of the function. Uh huh. Which is equal to the rate the A. D. T. Of pollutant put into the air given by this expression here. And we found that the result was about 713.654 standard pollution index.

Okay we are going to integrate our prime of T to get our of tea and then we'll be solving for our constant of integration given um the condition at our of one. So first of all to integrate E F T becomes E F T in the I direction. Are integral of Ellen. We need to do integration by parts. So we'll make our U R L N F T and R D V R T. So that means do you will be one over T and v will be T. So what we'll get here is a T. Ellen of tea. And then when we leave our table we have a negative D. T. To still integrate. So that will become a minus T. So that allows us to now put in R. J. Component for our empty and then as far as archaic component we are going to go up a power and divide by two so that two divided by two just becomes a one. Um then we have the plus C. And remember we had constants for R. I. J. And K. So when we're considering that when we place in a value of one for T we need to get out a value of zero in the I. Direction, one in the J. And one in the K. So here I've started to solve for my eyepiece, I'm gonna say zero equals E. To the first plus C. One. So C. One actually equals negative E. Now my next one is a value of one for the J. And then I'm also going to be putting in a value of one. So I have that negative one Ellen of one minus one. Now the Ellen of one is zero. So I really have a negative of a negative one there. And so when I subtract one from both sides I see that see to dis equals zero and then my third one. Um when I place a one in there, I also get a constant of zero. So the main thing is I need to have that minus E. In my eye component and then I can rewrite the rest of my R of T. And the vector differential has been assault.


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