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## Calculus I

by: Cristal Erdman

50

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10

# Calculus I MTH 201

Cristal Erdman
GVSU
GPA 3.58

William Dickinson

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COURSE
PROF.
William Dickinson
TYPE
Class Notes
PAGES
10
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 10 page Class Notes was uploaded by Cristal Erdman on Saturday September 26, 2015. The Class Notes belongs to MTH 201 at Grand Valley State University taught by William Dickinson in Fall. Since its upload, it has received 50 views. For similar materials see /class/214360/mth-201-grand-valley-state-university in Mathematics (M) at Grand Valley State University.

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Date Created: 09/26/15
Problem of the Fortnight 5 Brad Peirson Collaborated with Ryan Quaderer 11403 Will Dickinson MTH 201 G Math 201 Problem of the Fortnight 5 Review Deadline Thursday October 30 Due Tuesday November 4 Problem 5 A swimming pool is 60 feet long and 25 feet wide lts depth varies uniformly from 3 feet at the shallow end to 15 feet at the deep end as shown in the picture below I39iv l4 Suppose the pool has just been emptied and is now being filled with water at a rate of 200 cubic feet per minute a How long does it take to fill the entire pool with water b At what rate is the depth of water changing at the deep end when it is 5 feet deep at that end At what rate is the surface area of the water changing at the same instant Include several good pictures illustrating your calculations c At what rate is the depth of water changing at the deep end when it is 13 feet deep at that end At what rate is the surface area of the water changing at the same instant Include several good pictures illustrating your calculations d Is the depth at the deep end of the water increasing at a greater rate when the water is 8 feet deep or 10 feet deep Why Explain as fully as you possibly can a How long does it take to fill the entire pool with water In order to find the time required to fill the pool we first have to calculate the volume of the pool The diagram shows that the cross section of the pool is in the shape of a trapezoid The volume of the pool is therefore the area of this trapezoid times the width of the pool Vb1b2hgtltw Substituting the known dimensions of the pool from the diagram above we get V 15360gtlt 25 V 181500 V 91500 V 13500 f The total volume of the pool is 13500ft3 We are given the rate at which the pool is being filled so in order to find the time it will take to completely fill the pool all we need do is multiply the volume by the rate of filling 1min 200 3 675 min 13500ft3 gtlt At a fill rate of 200 cubic feet per minute the pool will be full in one hour seven minutes and thirty seconds B At what rate is the depth of water changing at the deep end when it is 5 feet deep at that end At what rate is the surface area of the water changing at the same instant First a diagram of the cross section of the pool in this situation will help in understanding the problem 60 12 The diagram shows that at a depth of twelve feet the cross section of the pool changes from trapezoidal to triangular in shape Therefore the 3 x 60 rectangular section on top of this triangle can be neglected in dealing with volumes at a depth less than or equal to twelve feet Since this cross section is triangular the volume of water at depths at or below twelve feet can be represented byV b 1225 Note that the 25 remains from the original volume equation because the width of the pool is the one dimension that remains constant The equation for the volume below twelve feet presents a new problem In order to find the rate of change of the depth of the water we will have to differentiate this new volume equation The problem arises in that as stated the volume equation holds three unknown variables So it becomes necessary to find a way to eliminate on of the variables Given the dimensions of the triangular portion of the pool we can find the angle at the bottom of the deep end 60 60 tant9 E tant9 5 t9 arctan5 t9 786901quot Knowing this angle we can now attempt to find a relationship between values for b and h The following table gives the value of h b tan786901 for all values of b up to 12 feet sjacoooxlcnuibwm xlu 0 01 This table of values of b and h show that for this triangular section of the pool h 5b This allows us to restate the volume equation Vlb5b25 2 V 25b2 25 V625b2 Now that our volume equation is down to two unknowns we can take the derivative with respect to time in order to find the rate of change in the height To do this it is helpful to think of both V and b as functions of time Vt and bt Vt 625bt2 db V39 t 125b I dt db It is necessary to include the termd because we are differentiating V with respect to t I and in order to do so we have to ag bt when we take its derivative In order to find db the rate of change in the height we have to now take this derivative and solve for I a V39 dt 125b The t term can be dropped so as to avoid confusion in the following steps V is constant as stated in the problem so In order to find the rate of change in the depth of the water at 5 feet all we now have to do is substitute 5 in for b dbi 200 57 db7200 57 db 32 t dt f min When the water is 5 feet deep in the deep end the depth is increasing by 32 feet per minute there The surface area can be modeled by the following diagram of the pool from above 25 X From the model we find that the surface area is equal to 25 x The rate of change in the surface area is then equal to the derivative of this equation A25x Also from above we know that the distance x is equal to 5 times the depth of water we are interested in A255b A125b Now we can differentiate this area equation to find the change in area with respect to time Agairt it is helpful for clarity s sake to consider A and b as functions of t At125bt db A t 125 dt We want to know the rate at which the surface area is changing when the water is five feet deep at the deep end of the pool Above we found that the rate of change in the depth of the water when it is five feet deep was 32 ft min Since we are interested in the rate of change of the surface area at a depth of five feet we can use this value for g in the above equation A t12532 A t 400 f min So the surface area of the water is increasing at a rate of forty square feet per minute when the water is five feet deep at the deep end C At what rate is the depth of water changing at the deep end when it is 13 feet deep at that end This problem is approached in much the same manner as part b The main difference is that now the depth is greater than twelve feet so the rectangular portion of the pools cross section must be taken into account 60 b2 b1 Therefore the cross section is again a trapezoid with area A 1511 b2 11 There is an added advantage in the trapezoidal model of the pool in that h is now a constant 60 feet So A 3011 b2 At this point the problem again shows up in the three unknowns Agairt we must try to relate b1 to b2 Using the geometry of the trapezoid the following table can be created showing values of b1 and the corresponding b2 values b1 b2 12 0 13 1 14 2 15 3 This table shows that at depths of water greater than 12 feet 2 b1712 It is also important to note that at a depth of twelve feet either model can be used The area of the cross section can now be expressed as A 3011 I1 712 A 30211 712 A 60bl 7 360 And since the width of the pool is still constant at 25 feet the volume can be expressed as V 601l 736025 V 15001l 7 9000 Agairt in order to take the derivative it is helpful to think of V and b1 as functions of time Vt 1 5001l 17 9000 V39 z 1 500 dt Again it is necessary to flag the derivative of b1 because we are differentiating volume With respect to time Now we can also drop the t term as above and solve for V39 dt 1500 V is still constant at 200 ftSmirt so 200 dt 1500 de 13 t dt f mm The depth is changing at rate of l ftmii1when the depth is 13 feet Indeed since there is no variable remaining in the derivative the derivative is constant At any depth over twelve feet the volume is increasing at a constant rate of liftmin The surface area can be modeled by the following diagram 25 60 The surface area of the water forms a rectangle of area A 2560 The rate of change of the surface area Will be equal to the derivative of this equation A39 0ft2 min The surface area is not increasing at a depth of 13 feet because both length and Width dimensions of the pool are constant here The lack of a variable in the derivative tells us that the rate of change is constant in this case it is zero We can also infer that the rate of change in the surface area Will be zero for all depths above 12 feet because at this depth the length of the water s surface becomes constant at 60 feet

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