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Exercise The temperature T in degrees Celsius of a chemical reaction is given as a function of time t, expressed in minutes, by the function defined on ¿ by: T (t )=(20 t +10)e−0.5t. 1) What is the initial temperature? 2) Show that T' (t )=(−10 t +15)e−0 .5t. 3) Study the sign of T' (t ), then draw up the table of variations of T . We do not ask for the limit of T in +∞. 4) What is the maximum temperature reached by the reaction chemical. We will give an approximate value to within 10−2. 5) After how long does the temperature T go back down to its initial value? We will give an approximate value of this time in minutes and seconds. DM 2: study of a function Exercise The temperature T in degrees Celsius of a chemical reaction is given as a function of time t, expressed in minutes, by the function defined on ¿ by: T (t )=(20 t +10)e−0.5t. 1) What is the initial temperature? 2) Show that T' (t )=(−10 t +15)e−0.5 t. 3) Study the sign of T' (t ), then draw up the table of variations of T . We do not ask for the limit of T in +∞. 4) What is the maximum temperature reached by the reaction chemical. We will give an approximate value to within 10−2. 5) After how long does the temperature T go back down to its initial value? We will give an approximate value of this time in minutes and seconds.

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Answer to a math question Exercise The temperature T in degrees Celsius of a chemical reaction is given as a function of time t, expressed in minutes, by the function defined on ¿ by: T (t )=(20 t +10)e−0.5t. 1) What is the initial temperature? 2) Show that T' (t )=(−10 t +15)e−0 .5t. 3) Study the sign of T' (t ), then draw up the table of variations of T . We do not ask for the limit of T in +∞. 4) What is the maximum temperature reached by the reaction chemical. We will give an approximate value to within 10−2. 5) After how long does the temperature T go back down to its initial value? We will give an approximate value of this time in minutes and seconds. DM 2: study of a function Exercise The temperature T in degrees Celsius of a chemical reaction is given as a function of time t, expressed in minutes, by the function defined on ¿ by: T (t )=(20 t +10)e−0.5t. 1) What is the initial temperature? 2) Show that T' (t )=(−10 t +15)e−0.5 t. 3) Study the sign of T' (t ), then draw up the table of variations of T . We do not ask for the limit of T in +∞. 4) What is the maximum temperature reached by the reaction chemical. We will give an approximate value to within 10−2. 5) After how long does the temperature T go back down to its initial value? We will give an approximate value of this time in minutes and seconds.

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Velda
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1) La température initiale est la valeur de T lorsque t = 0. Pour la trouver, il suffit de brancher 0 pour t dans la fonction : T(0) = (20 \times 0 + 10)e^{ -0,5 \times 0} = 10e^0 = 10 La température initiale est donc de 10°C. 2) Pour montrer que T'(t) = (-10t + 15)e^(-0.5t), nous devons utiliser la règle du produit et la règle de la chaîne de différenciation. La règle du produit dit que si f et g sont deux fonctions, alors (f \times g)' = f' \times g + f \times g'. La règle de la chaîne dit que si h est une fonction de g , et g est une fonction de x, alors (h \circ g)'(x) = h'(g(x)) \times g'(x). Dans ce cas, on peut écrire T comme un produit de deux fonctions : T(t) = f(t) \times g(t)f(t) = 20t + 10 et g(t) = e^{- 0.5t} Ensuite, en utilisant la règle du produit, on obtient : T'(t) = f'(t) \times g(t) + f(t) \times g'(t) Pour trouver f'(t) et g'(t), nous devons utiliser la règle de la chaîne. Pour f'(t), on a : f'(t) = \frac{d}{dt}(20t + 10) = 20 + \frac{d}{dt}(10) = 20 + 0 = 20 Pour g'(t), on a : g'(t) = \frac{d}{dt}(e^{-0.5t}) = e^{-0.5t} \times \frac {d}{dt}(-0.5t) = e^{-0.5t} \times (-0.5) = -0.5e^{-0.5t} En branchant ces valeurs dans la règle du produit, nous obtenons : T'(t) = 20 \times e^{-0.5t} + (20t + 10) \times (-0.5e^{-0.5t}) En simplifiant, on obtient : T'(t) = (20 - 10t - 5)e^{-0.5t} En factorisant -5, nous obtenons : T'(t) = -5(2 - 2t - 1)e^{-0.5t} En simplifiant davantage, nous obtenons : T'(t) = (-10t + 15)e^{-0.5t} C'est la même chose que l'expression donnée, nous avons donc montré que T'(t) = ( -10t + 15)e^(-0,5t). 3) Pour étudier le signe de T'(t), nous devons trouver les valeurs de t qui rendent T'(t) égal à zéro ou indéfini. Puisque T'(t) est une fonction continue, elle n'est jamais indéfinie. Pour trouver les zéros de T'(t), nous devons résoudre l'équation : (-10t + 15)e^{-0,5t} = 0 Cette équation n'a qu'une seule solution, qui est t = 1,5. Cela signifie que T'(t) change de signe à t = 1,5. Pour trouver le signe de T'(t) sur chaque intervalle, on peut utiliser un point test. Par exemple, pour t < 1,5, nous pouvons utiliser t = 0 et le brancher sur T'(t) : T'(0) = (-10 \times 0 + 15)e^{-0,5 \times 0} = 15e^0 = 15 Puisque c'est positif, T'(t) est positif pour t < 1,5. De même, pour t > 1,5, nous pouvons utiliser t = 2 et le brancher sur T'(t) : T'(2) = (-10 \times 2 + 15)e^{-0,5 \times 2} = -5e^{-1} Puisque c'est négatif, T'(t) est négatif pour t > 1,5. On peut donc tracer le tableau des variations de T comme suit : | t | -∞ | 1.5 | +∞ | | T'(t) | + | 0 | - | | T(t) | ↗ | maximum | ↘ | 4) La température maximale atteinte par la réaction chimique est la valeur de T à t = 1,5, qui est le point où T'(t) passe du positif au négatif. Pour le trouver, il suffit de brancher 1.5 pour t dans la fonction : T(1.5) = (20 \times 1.5 + 10)e^{-0.5 \times 1.5} = 40e^{-0.75} À l'aide d'une calculatrice, nous obtenons : T(1,5) \environ 18,89 Par conséquent, la température maximale atteinte par la réaction chimique est de 18,89°C (à 10^-2^ près). 5) La température T redescend à sa valeur initiale lorsque T(t) = 10. Pour trouver l'instant où cela se produit, il faut résoudre l'équation : (20t + 10)e^{-0.5t} = 10 En divisant les deux côtés par 10, on obtient : (2t + 1)e^{-0.5t} = 1 En prenant le logarithme népérien des deux côtés, on obtient : \ln((2t + 1 )e^{-0.5t}) = \ln(1) En utilisant les propriétés des logarithmes, on obtient : \ln(2t + 1) - 0.5t = 0 Cette équation ne peut pas être résolue algébriquement, donc nous devons utiliser une méthode numérique, telle qu'une calculatrice graphique ou un solveur en ligne, pour trouver une solution approximative. Une de ces solutions est : t \environ 4,67 Par conséquent, la température T redescend à sa valeur initiale après 4,67 minutes. Pour convertir cela en minutes et secondes, il faut multiplier la partie décimale par 60 : 0,67 \times 60 \approx 40,3992 La température T redescend donc à sa valeur initiale après 4 minutes et 40 secondes.

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