My question is with a multi-part homework assignment that we got in class. The information given is :
A factory manufactures widgets. The rate of production of widgets after t weeks is widgets/ week. The equation for this is dx/dt=30(1-(20/((t+20)^2))).
The actual questions are below.
1)Determine the rate at which widgets are produced at the start of week 1. Set up the expression but do not solve.
For this one I just plugged "1" into the given equation. So the number of widgets produced at the start of week 1 is equal to 30(1-(20/(((1)+20)^2))). I'm pretty sure I am correct in this.
2)Determine the number of widgets produced from the beginning of production to the beginning of the fifth week.
For this I set up a definite integral of the given equation with b=5 and a=0. I am unsure whether I should put the given equation into the definite integral or if because the given equation is a rate, if I need to first take the anti-derivative of the given equation and then put that into the definite integral from 0 to 5. This question is where I am unsure to proceed.
3)Determine the number of widgets produced from the beginning of the fifth week to the end of the ninth week.
I thought to set up a definite integral from 5 to 10 (10, because it wants to the end of the ninth week). As I said above I am also not sure if I need to first take the anti-derivative of the given equation and then put that into the definite integral. I am also unsure about where to proceed with this question.
For this one I just plugged "1" into the given equation. So the number of widgets produced at the start of week 1 is equal to 30(1-(20/(((1)+20)^2))). I'm pretty sure I am correct in this>>Nope. for the start of week 1, use t=0. Reread what the dn/dt represents (END OF WEEK T)
<<given equation into the definite integral or if because the given equation is a rate, if I need to first take the anti-derivative of the given equation and then put that into the definite integral from 0 to 5. This question is where I am unsure to proceed.>> b should be 4, same reason as above.
On 3, should be 4 to 9.
So take the integral of the function. Then apply the limits correctly
To solve question 1, you correctly plugged in 1 into the given equation to find the rate at which widgets are produced at the start of week 1. The expression you set up is 30(1-(20/((1+20)^2))). While you didn't simplify the expression further, you have correctly set it up.
For question 2, you want to determine the number of widgets produced from the beginning of production to the beginning of the fifth week. To do this, you need to integrate the given equation over the interval from 0 to 5. Since the given equation dx/dt represents the rate of production, you do not need to take any additional steps such as finding the antiderivative. You can directly integrate the given equation with respect to t as follows:
∫[0,5] 30(1-(20/((t+20)^2))) dt
To find the solution to this integral, you can use any integration technique or software tool you prefer.
For question 3, you want to determine the number of widgets produced from the beginning of the fifth week to the end of the ninth week. Similar to question 2, you can set up a definite integral to find this value. Since you're looking for the number of widgets produced during the time interval from 5 to 9 (end of the ninth week), you can set up the following definite integral:
∫[5,9] 30(1-(20/((t+20)^2))) dt
Again, you do not need to take the antiderivative of the given equation since it already represents the rate of production. Just evaluate the integral using your preferred integration method or software tool to find the answer.
Note: Remember to evaluate the definite integrals using the limits of integration, in this case, [0,5] and [5,9].