What is the total cost to heat the greenhouse during this 24-hour period?












2












$begingroup$


The temperature outside the greenhouse during a $24$-hour period is given by $F(t) = 22 + 20cos(frac{pi t}{12}), 0 le t le 24$, where $F(t)$ is measured in degrees Fahrenheit and t is measured in hours. Let $t=0$ represent noon on the first day.



d. The heater in the greenhouse kicks on whenever the outdoor temperature is at or below $36$ degrees Fahrenheit. For what values of $t$ is the heater on?



I used a graphing calculator for this part, between $[3.082, 20.962]$.



e. The cost of heating the greenhouse accumulates at a rate of $$0.06$ per hour for each degree the outside temperature falls below $36$ degrees Fahrenheit. What is the total cost to heat the greenhouse during this $24$-hour period?



I'm not really understanding where to start on this problem. My teacher suggested finding the integral difference between $y=36$ and $F(t)$ when the degrees fall under $36$, but I'm confused on what this represents and what to do with it. If anyone could help me understand, that would be much appreciated.



Thank you!






calculus







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$endgroup$












  • $begingroup$
    You can make your integral easier by just considering when $cos{frac{pi t}{12}}$ is not greater than $frac{7}{10}$.
    $endgroup$
    – John Douma
    Dec 24 '18 at 4:55










  • $begingroup$
    You have a typo : $3.082$ should be $3.0382$
    $endgroup$
    – Claude Leibovici
    Dec 24 '18 at 10:44
















2












$begingroup$


The temperature outside the greenhouse during a $24$-hour period is given by $F(t) = 22 + 20cos(frac{pi t}{12}), 0 le t le 24$, where $F(t)$ is measured in degrees Fahrenheit and t is measured in hours. Let $t=0$ represent noon on the first day.



d. The heater in the greenhouse kicks on whenever the outdoor temperature is at or below $36$ degrees Fahrenheit. For what values of $t$ is the heater on?



I used a graphing calculator for this part, between $[3.082, 20.962]$.



e. The cost of heating the greenhouse accumulates at a rate of $$0.06$ per hour for each degree the outside temperature falls below $36$ degrees Fahrenheit. What is the total cost to heat the greenhouse during this $24$-hour period?



I'm not really understanding where to start on this problem. My teacher suggested finding the integral difference between $y=36$ and $F(t)$ when the degrees fall under $36$, but I'm confused on what this represents and what to do with it. If anyone could help me understand, that would be much appreciated.



Thank you!






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can make your integral easier by just considering when $cos{frac{pi t}{12}}$ is not greater than $frac{7}{10}$.
    $endgroup$
    – John Douma
    Dec 24 '18 at 4:55










  • $begingroup$
    You have a typo : $3.082$ should be $3.0382$
    $endgroup$
    – Claude Leibovici
    Dec 24 '18 at 10:44














2












2








2


1



$begingroup$


The temperature outside the greenhouse during a $24$-hour period is given by $F(t) = 22 + 20cos(frac{pi t}{12}), 0 le t le 24$, where $F(t)$ is measured in degrees Fahrenheit and t is measured in hours. Let $t=0$ represent noon on the first day.



d. The heater in the greenhouse kicks on whenever the outdoor temperature is at or below $36$ degrees Fahrenheit. For what values of $t$ is the heater on?



I used a graphing calculator for this part, between $[3.082, 20.962]$.



e. The cost of heating the greenhouse accumulates at a rate of $$0.06$ per hour for each degree the outside temperature falls below $36$ degrees Fahrenheit. What is the total cost to heat the greenhouse during this $24$-hour period?



I'm not really understanding where to start on this problem. My teacher suggested finding the integral difference between $y=36$ and $F(t)$ when the degrees fall under $36$, but I'm confused on what this represents and what to do with it. If anyone could help me understand, that would be much appreciated.



Thank you!






calculus







share|cite|improve this question











$endgroup$




The temperature outside the greenhouse during a $24$-hour period is given by $F(t) = 22 + 20cos(frac{pi t}{12}), 0 le t le 24$, where $F(t)$ is measured in degrees Fahrenheit and t is measured in hours. Let $t=0$ represent noon on the first day.



d. The heater in the greenhouse kicks on whenever the outdoor temperature is at or below $36$ degrees Fahrenheit. For what values of $t$ is the heater on?



I used a graphing calculator for this part, between $[3.082, 20.962]$.



e. The cost of heating the greenhouse accumulates at a rate of $$0.06$ per hour for each degree the outside temperature falls below $36$ degrees Fahrenheit. What is the total cost to heat the greenhouse during this $24$-hour period?



I'm not really understanding where to start on this problem. My teacher suggested finding the integral difference between $y=36$ and $F(t)$ when the degrees fall under $36$, but I'm confused on what this represents and what to do with it. If anyone could help me understand, that would be much appreciated.



Thank you!






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 4:05









Kemono Chen

2,9801740




2,9801740










asked Dec 24 '18 at 4:01









Ran HanRan Han

142




142












  • $begingroup$
    You can make your integral easier by just considering when $cos{frac{pi t}{12}}$ is not greater than $frac{7}{10}$.
    $endgroup$
    – John Douma
    Dec 24 '18 at 4:55










  • $begingroup$
    You have a typo : $3.082$ should be $3.0382$
    $endgroup$
    – Claude Leibovici
    Dec 24 '18 at 10:44


















  • $begingroup$
    You can make your integral easier by just considering when $cos{frac{pi t}{12}}$ is not greater than $frac{7}{10}$.
    $endgroup$
    – John Douma
    Dec 24 '18 at 4:55










  • $begingroup$
    You have a typo : $3.082$ should be $3.0382$
    $endgroup$
    – Claude Leibovici
    Dec 24 '18 at 10:44
















$begingroup$
You can make your integral easier by just considering when $cos{frac{pi t}{12}}$ is not greater than $frac{7}{10}$.
$endgroup$
– John Douma
Dec 24 '18 at 4:55




$begingroup$
You can make your integral easier by just considering when $cos{frac{pi t}{12}}$ is not greater than $frac{7}{10}$.
$endgroup$
– John Douma
Dec 24 '18 at 4:55












$begingroup$
You have a typo : $3.082$ should be $3.0382$
$endgroup$
– Claude Leibovici
Dec 24 '18 at 10:44




$begingroup$
You have a typo : $3.082$ should be $3.0382$
$endgroup$
– Claude Leibovici
Dec 24 '18 at 10:44










3 Answers
3






active

oldest

votes


















4












$begingroup$

The temperature outside is $$F(t)=22+20cos(frac{pi t}{12})$$ The temperature difference if $Delta T(t)=36-F(t)$, but you only need to consider it if $F(t)lt 36$, or $Delta T>0$. Let's consider a small interval of time $d t$. Since the temperature difference is $Delta T$, the price for this interval is $$0.06Delta T(t) dt$$ You need to add together the prices for all these small time intervals, which means doing the integral.
$$mathrm{Cost}=int_{t_1}^{t_2}0.06(36-22-20cos(frac{pi t}{12}))dt$$
Here $t_1$ is when the temperature dips below 36 degrees, and $t_2$ is when it raises to 36 again.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like your explanation!
    $endgroup$
    – Larry
    Dec 24 '18 at 4:51



















1












$begingroup$

This question is similar with question 5 from the 1998 Calc AB FRQ . Only the numbers are changed.
So you have found $$tin[3.082,20.962]$$
Then the setup for part e. will be the following:
$$begin{align}
C_{total~cost} &= 0.06int_{3.082}^{20.962}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt\
&=0.06(360.03979)=21.602approx$21.6
end{align}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Just for the fun of it.



    You used a graphing calculator for solving
    $$22 + 20cos(frac{pi t}{12})=36 implies cos(frac{pi t}{12})=frac 7{10}$$ You could have used the approximation
    $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$ making $x=frac{pi t}{12}$, we end then with
    $$frac{144-4 t^2 }{144+t^2}=frac 7{10}implies 47t^2=432implies t=12 sqrt{frac{3}{47}}approx 3.03175$$



    $$C_{total~cost} = frac 6 {100} int_{12 sqrt{frac{3}{47}}}^{24-12 sqrt{frac{3}{47}}}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt$$ making the nice
    $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{144 }{5 pi }sin left(sqrt{frac{3}{47}} pi right)$$ We can continue using the magnificent approximation
    $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago)
    $$sin left(sqrt{frac{3}{47}} pi right)simeq frac{16 left(235 sqrt{141}-177right)}{58753}$$
    $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{2304 left(235 sqrt{141}-177right)}{293765 pi }approx 21.5912$$ while a completely rigorous calculation would give
    $$C_{total~cost}=frac{72 left(sqrt{51}+7 pi -7 cos ^{-1}left(frac{7}{10}right)right)}{25
    pi }approx 21.6026$$



    I hope and wish that we shall not argue for one cent difference.



    Merry Xmas






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      The temperature outside is $$F(t)=22+20cos(frac{pi t}{12})$$ The temperature difference if $Delta T(t)=36-F(t)$, but you only need to consider it if $F(t)lt 36$, or $Delta T>0$. Let's consider a small interval of time $d t$. Since the temperature difference is $Delta T$, the price for this interval is $$0.06Delta T(t) dt$$ You need to add together the prices for all these small time intervals, which means doing the integral.
      $$mathrm{Cost}=int_{t_1}^{t_2}0.06(36-22-20cos(frac{pi t}{12}))dt$$
      Here $t_1$ is when the temperature dips below 36 degrees, and $t_2$ is when it raises to 36 again.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I like your explanation!
        $endgroup$
        – Larry
        Dec 24 '18 at 4:51
















      4












      $begingroup$

      The temperature outside is $$F(t)=22+20cos(frac{pi t}{12})$$ The temperature difference if $Delta T(t)=36-F(t)$, but you only need to consider it if $F(t)lt 36$, or $Delta T>0$. Let's consider a small interval of time $d t$. Since the temperature difference is $Delta T$, the price for this interval is $$0.06Delta T(t) dt$$ You need to add together the prices for all these small time intervals, which means doing the integral.
      $$mathrm{Cost}=int_{t_1}^{t_2}0.06(36-22-20cos(frac{pi t}{12}))dt$$
      Here $t_1$ is when the temperature dips below 36 degrees, and $t_2$ is when it raises to 36 again.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I like your explanation!
        $endgroup$
        – Larry
        Dec 24 '18 at 4:51














      4












      4








      4





      $begingroup$

      The temperature outside is $$F(t)=22+20cos(frac{pi t}{12})$$ The temperature difference if $Delta T(t)=36-F(t)$, but you only need to consider it if $F(t)lt 36$, or $Delta T>0$. Let's consider a small interval of time $d t$. Since the temperature difference is $Delta T$, the price for this interval is $$0.06Delta T(t) dt$$ You need to add together the prices for all these small time intervals, which means doing the integral.
      $$mathrm{Cost}=int_{t_1}^{t_2}0.06(36-22-20cos(frac{pi t}{12}))dt$$
      Here $t_1$ is when the temperature dips below 36 degrees, and $t_2$ is when it raises to 36 again.






      share|cite|improve this answer









      $endgroup$



      The temperature outside is $$F(t)=22+20cos(frac{pi t}{12})$$ The temperature difference if $Delta T(t)=36-F(t)$, but you only need to consider it if $F(t)lt 36$, or $Delta T>0$. Let's consider a small interval of time $d t$. Since the temperature difference is $Delta T$, the price for this interval is $$0.06Delta T(t) dt$$ You need to add together the prices for all these small time intervals, which means doing the integral.
      $$mathrm{Cost}=int_{t_1}^{t_2}0.06(36-22-20cos(frac{pi t}{12}))dt$$
      Here $t_1$ is when the temperature dips below 36 degrees, and $t_2$ is when it raises to 36 again.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 24 '18 at 4:45









      AndreiAndrei

      11.7k21026




      11.7k21026












      • $begingroup$
        I like your explanation!
        $endgroup$
        – Larry
        Dec 24 '18 at 4:51


















      • $begingroup$
        I like your explanation!
        $endgroup$
        – Larry
        Dec 24 '18 at 4:51
















      $begingroup$
      I like your explanation!
      $endgroup$
      – Larry
      Dec 24 '18 at 4:51




      $begingroup$
      I like your explanation!
      $endgroup$
      – Larry
      Dec 24 '18 at 4:51











      1












      $begingroup$

      This question is similar with question 5 from the 1998 Calc AB FRQ . Only the numbers are changed.
      So you have found $$tin[3.082,20.962]$$
      Then the setup for part e. will be the following:
      $$begin{align}
      C_{total~cost} &= 0.06int_{3.082}^{20.962}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt\
      &=0.06(360.03979)=21.602approx$21.6
      end{align}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        This question is similar with question 5 from the 1998 Calc AB FRQ . Only the numbers are changed.
        So you have found $$tin[3.082,20.962]$$
        Then the setup for part e. will be the following:
        $$begin{align}
        C_{total~cost} &= 0.06int_{3.082}^{20.962}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt\
        &=0.06(360.03979)=21.602approx$21.6
        end{align}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          This question is similar with question 5 from the 1998 Calc AB FRQ . Only the numbers are changed.
          So you have found $$tin[3.082,20.962]$$
          Then the setup for part e. will be the following:
          $$begin{align}
          C_{total~cost} &= 0.06int_{3.082}^{20.962}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt\
          &=0.06(360.03979)=21.602approx$21.6
          end{align}$$






          share|cite|improve this answer









          $endgroup$



          This question is similar with question 5 from the 1998 Calc AB FRQ . Only the numbers are changed.
          So you have found $$tin[3.082,20.962]$$
          Then the setup for part e. will be the following:
          $$begin{align}
          C_{total~cost} &= 0.06int_{3.082}^{20.962}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt\
          &=0.06(360.03979)=21.602approx$21.6
          end{align}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 4:34









          LarryLarry

          2,3042828




          2,3042828























              1












              $begingroup$

              Just for the fun of it.



              You used a graphing calculator for solving
              $$22 + 20cos(frac{pi t}{12})=36 implies cos(frac{pi t}{12})=frac 7{10}$$ You could have used the approximation
              $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$ making $x=frac{pi t}{12}$, we end then with
              $$frac{144-4 t^2 }{144+t^2}=frac 7{10}implies 47t^2=432implies t=12 sqrt{frac{3}{47}}approx 3.03175$$



              $$C_{total~cost} = frac 6 {100} int_{12 sqrt{frac{3}{47}}}^{24-12 sqrt{frac{3}{47}}}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt$$ making the nice
              $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{144 }{5 pi }sin left(sqrt{frac{3}{47}} pi right)$$ We can continue using the magnificent approximation
              $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago)
              $$sin left(sqrt{frac{3}{47}} pi right)simeq frac{16 left(235 sqrt{141}-177right)}{58753}$$
              $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{2304 left(235 sqrt{141}-177right)}{293765 pi }approx 21.5912$$ while a completely rigorous calculation would give
              $$C_{total~cost}=frac{72 left(sqrt{51}+7 pi -7 cos ^{-1}left(frac{7}{10}right)right)}{25
              pi }approx 21.6026$$



              I hope and wish that we shall not argue for one cent difference.



              Merry Xmas






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Just for the fun of it.



                You used a graphing calculator for solving
                $$22 + 20cos(frac{pi t}{12})=36 implies cos(frac{pi t}{12})=frac 7{10}$$ You could have used the approximation
                $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$ making $x=frac{pi t}{12}$, we end then with
                $$frac{144-4 t^2 }{144+t^2}=frac 7{10}implies 47t^2=432implies t=12 sqrt{frac{3}{47}}approx 3.03175$$



                $$C_{total~cost} = frac 6 {100} int_{12 sqrt{frac{3}{47}}}^{24-12 sqrt{frac{3}{47}}}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt$$ making the nice
                $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{144 }{5 pi }sin left(sqrt{frac{3}{47}} pi right)$$ We can continue using the magnificent approximation
                $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago)
                $$sin left(sqrt{frac{3}{47}} pi right)simeq frac{16 left(235 sqrt{141}-177right)}{58753}$$
                $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{2304 left(235 sqrt{141}-177right)}{293765 pi }approx 21.5912$$ while a completely rigorous calculation would give
                $$C_{total~cost}=frac{72 left(sqrt{51}+7 pi -7 cos ^{-1}left(frac{7}{10}right)right)}{25
                pi }approx 21.6026$$



                I hope and wish that we shall not argue for one cent difference.



                Merry Xmas






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Just for the fun of it.



                  You used a graphing calculator for solving
                  $$22 + 20cos(frac{pi t}{12})=36 implies cos(frac{pi t}{12})=frac 7{10}$$ You could have used the approximation
                  $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$ making $x=frac{pi t}{12}$, we end then with
                  $$frac{144-4 t^2 }{144+t^2}=frac 7{10}implies 47t^2=432implies t=12 sqrt{frac{3}{47}}approx 3.03175$$



                  $$C_{total~cost} = frac 6 {100} int_{12 sqrt{frac{3}{47}}}^{24-12 sqrt{frac{3}{47}}}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt$$ making the nice
                  $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{144 }{5 pi }sin left(sqrt{frac{3}{47}} pi right)$$ We can continue using the magnificent approximation
                  $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago)
                  $$sin left(sqrt{frac{3}{47}} pi right)simeq frac{16 left(235 sqrt{141}-177right)}{58753}$$
                  $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{2304 left(235 sqrt{141}-177right)}{293765 pi }approx 21.5912$$ while a completely rigorous calculation would give
                  $$C_{total~cost}=frac{72 left(sqrt{51}+7 pi -7 cos ^{-1}left(frac{7}{10}right)right)}{25
                  pi }approx 21.6026$$



                  I hope and wish that we shall not argue for one cent difference.



                  Merry Xmas






                  share|cite|improve this answer











                  $endgroup$



                  Just for the fun of it.



                  You used a graphing calculator for solving
                  $$22 + 20cos(frac{pi t}{12})=36 implies cos(frac{pi t}{12})=frac 7{10}$$ You could have used the approximation
                  $$cos(x) simeqfrac{pi ^2-4x^2}{pi ^2+x^2}qquad (-frac pi 2 leq xleqfrac pi 2)$$ making $x=frac{pi t}{12}$, we end then with
                  $$frac{144-4 t^2 }{144+t^2}=frac 7{10}implies 47t^2=432implies t=12 sqrt{frac{3}{47}}approx 3.03175$$



                  $$C_{total~cost} = frac 6 {100} int_{12 sqrt{frac{3}{47}}}^{24-12 sqrt{frac{3}{47}}}left(36-left[22+20cosleft(frac{pi t}{12}right)right]right)dt$$ making the nice
                  $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{144 }{5 pi }sin left(sqrt{frac{3}{47}} pi right)$$ We can continue using the magnificent approximation
                  $$sin(x) simeq frac{16 (pi -x) x}{5 pi ^2-4 (pi -x) x}qquad (0leq xleqpi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1400$ years ago)
                  $$sin left(sqrt{frac{3}{47}} pi right)simeq frac{16 left(235 sqrt{141}-177right)}{58753}$$
                  $$C_{total~cost}=frac{504 left(47-sqrt{141}right)}{1175}+frac{2304 left(235 sqrt{141}-177right)}{293765 pi }approx 21.5912$$ while a completely rigorous calculation would give
                  $$C_{total~cost}=frac{72 left(sqrt{51}+7 pi -7 cos ^{-1}left(frac{7}{10}right)right)}{25
                  pi }approx 21.6026$$



                  I hope and wish that we shall not argue for one cent difference.



                  Merry Xmas







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 24 '18 at 11:00

























                  answered Dec 24 '18 at 10:45









                  Claude LeiboviciClaude Leibovici

                  120k1157132




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