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.

141

likes
705 views

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.

Expert avatar
Velda
4.5
109 Answers
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.

Frequently asked questions (FAQs)
Math Question: Graph the inequality y > 2x + 1.
+
Math question: "Factorize the expression 5x^2 + 7xy - 6y^2 using the factoring formulas."
+
What is the equation of an ellipse with the major axis of length 10 and the minor axis of length 6?
+
New questions in Mathematics
solve the following trigo equation for 0°<= x <= 360°. sec x =-2
The profit G of the company CHUNCHES SA is given by G(x) = 3×(40 – ×), where × is the quantity of items sold. Find the maximum profit.
2.5 / 21.85
An electrical company manufactures batteries that have a duration that is distributed approximately normally, with a mean of 700 hours and a standard deviation of 40 hours. Find the probability that a randomly selected battery has an average life of less than 810 hours.
the probabilty that a person has a motorcycle, given that she owns a car 25%. the percentage of people owing a motorcycle is 15% and that who own a car is 35%. find probabilty that a person owns any one or both of those
How many anagrams of the word SROMEC there that do not contain STROM, MOST, MOC or CEST as a subword? By subword is meant anything that is created by omitting some letters - for example, the word EMROSCT contains both MOC and MOST as subwords.
What is the appropriate measurement for the weight of an African elephant?
Log5 625
. What will be the osmotic pressure of a solution that was prepared at 91°F by dissolving 534 grams of aluminum hydroxide in enough water to generate 2.784 ml of solution.
show step by step simplification: (¬𝑑∨((¬b∧c)∨(b∧¬c)))∧((𝑎 ∧ 𝑏) ∨ (¬𝑎 ∧ ¬𝑏))∧(¬𝑐∨((¬𝑑∧𝑎)∨(𝑑∧¬𝑎)))
A box of numbered pens has 12 red, 12 blue, 12 green and 12 yellow pens. The pens for each colour are numbered from 1 to 12. There is a unique number on each pen, so no pen is exactly the same as any other pen in the box. When reaching into the box to randomly draw five pens without replacement, what is the proportion of getting exactly four pens of the same colour (Note: the numbers matter but the order does not)?
2)A tourist has 15 pairs of pants in his hotel room closet. Suppose 5 are blue and the rest are black. The tourist leaves his room twice a day. He takes a pair of pants and puts them on, the tourist leaves the first pair of pants in the closet again and takes another one and puts them on. What is the probability that the two pants chosen are black?
Fill in the P(X-x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -5 ,3 , 4, 5 , and 6.
When Sara was 15 years old, an uncle left her as inheritanceà a sum of 10,000 euros which he invested in a bank that applies the interest rate of 2,5% annual. Today Sara is 18 years and wants to buy a'car, how much she can ò withdraw from the bank?
A company made 150,000 in the first year 145,000 in the second 140,000 in the third year successively during the first decade of this company's existence it made a total of
Let X be a discrete random variable such that E(X)=3 and V(X)=5. Let 𝑌 = 2𝑋^2 − 3𝑋. Determine E(Y).
Find the area of a triangle ABC when m<C = 14 degrees, a = 5.7 miles, and b = 9.3 miles.
For the numbers below, select a number at random and find the probability that: a. The number is even b. The sum of the number’s digit is even c. The number is greater than 50 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
How much does 7.2 moles of ammonium dichromate weigh? (NH4)2Cr2O7
The perimeter of a rectangular rug is 42 feet. The width is 9 feet. What is the length?