Question Video: Finding the Rate of Change of a Cubic Function That Represents Production in a Chemical Reaction | Nagwa Question Video: Finding the Rate of Change of a Cubic Function That Represents Production in a Chemical Reaction | Nagwa

Question Video: Finding the Rate of Change of a Cubic Function That Represents Production in a Chemical Reaction Mathematics • Second Year of Secondary School

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The output in mg of a chemical reaction after 𝑡 seconds is given by 𝑦 = 4𝑡³. What is the instantaneous rate of production of this reaction at 𝑡 = 2 seconds?

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Video Transcript

The output in milligrams of a chemical reaction after 𝑡 seconds is given by 𝑦 is equal to four 𝑡 cubed. What is the instantaneous rate of production of this reaction at 𝑡 equals two seconds?

In this question, we are given an equation for the output 𝑦 in milligrams of a chemical reaction in terms of time 𝑡 seconds. We will begin by rewriting this as a function where 𝑦 is equal to 𝑓 of 𝑡, and this is equal to four 𝑡 cubed. We are asked to find the instantaneous rate of change of this function at 𝑡 equals two seconds. We recall that the rate of change of a function 𝑓 at 𝑎 is given by 𝑓 prime evaluated at 𝑎. And this is equal to the limit as ℎ tends to zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all divided by ℎ.

As we want to calculate the rate of change at 𝑡 equals two seconds, we will substitute 𝑎 equal to two. We will now substitute 𝑡 is equal to two plus ℎ and 𝑡 is equal to two into our function. 𝑓 of two plus ℎ is equal to four multiplied by two plus ℎ all cubed. Using our knowledge of a binomial expansion where 𝑎 plus 𝑏 all cubed is equal to 𝑎 cubed plus three 𝑎 squared 𝑏 plus three 𝑎𝑏 squared plus 𝑏 cubed, then cubing two plus ℎ gives us eight plus 12ℎ plus six ℎ squared plus ℎ cubed. We can then multiply through by four such that 𝑓 of two plus ℎ is equal to 32 plus 48ℎ plus 24ℎ squared plus four ℎ cubed.

𝑓 of two is equal to four multiplied by two cubed, which is equal to 32. Substituting our expressions for 𝑓 of two plus ℎ and 𝑓 of two, we have 𝑓 prime of two is equal to the limit as ℎ tends to zero of 32 plus 48ℎ plus 24ℎ squared plus four ℎ cubed minus 32 all divided by ℎ. 32 minus 32 is equal to zero. We can then divide through by ℎ, giving us the limit as ℎ tends to zero of 48 plus 24ℎ plus four ℎ squared. We now have a polynomial in ℎ, and we can attempt to solve this using direct substitution. Replacing ℎ with zero, we have 48 plus 24 multiplied by zero plus four multiplied by zero squared, which is equal to 48.

The instantaneous rate of change is therefore equal to 48. And since the output was measured in milligrams and the time in seconds, we have a rate of production equal to 48 milligrams per second.

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