How could I solve c) and d) without the use of derivatives for the following question?
It is estimated that, t years from now, the amount of waste accumulated, Q, in tonnes, will be Q(t)= 10^4(t^2+15t+70) tonnes, t is greater than or equal to 0, t is also less than or equal to 10.
b) What will the average rate of change of this quauntity over the next three years?
c) What is the present rate of change of this quantity?
d) When will the rate of change reach 3.0 * 10^5 tonnes per year?
This little essay should help.
http://jwilson.coe.uga.edu/EMAT6680Fa2012/Faircloth/Essay4alf/TangentLine.html
To solve parts c) and d) without using derivatives, we can estimate the present rate of change and the time when the rate of change reaches a certain value by using average rates of change.
For part c), the present rate of change is the average rate of change of the quantity over an infinitesimally small interval of time. We can estimate it by finding the average rate of change over a small interval of time, such as 0 to 1 year.
To determine the average rate of change over the interval (0, 1), we need to find the quantity at the beginning of the interval, Q(0), and at the end of the interval, Q(1).
Substituting t=0 into the equation Q(t) = 10^4(t^2 + 15t + 70), we find Q(0) = 10^4(0^2 + 15(0) + 70) = 70,000 tonnes.
Substituting t=1 into the equation Q(t) = 10^4(t^2 + 15t + 70), we find Q(1) = 10^4(1^2 + 15(1) + 70) = 96,000 tonnes.
Therefore, the average rate of change over the interval (0, 1) is:
Average rate of change = (Q(1) - Q(0)) / (1 - 0) = (96,000 - 70,000) / 1 = 26,000 tonnes per year.
For part d), we are given that the rate of change is 3.0 * 10^5 tonnes per year. To find when this rate of change occurs, we need to estimate the time it takes for the average rate of change to reach this value.
We can set up an equation using the average rate of change and solve for t:
Average rate of change = (Q(t) - Q(0)) / (t - 0)
Substituting the average rate of change of 3.0 * 10^5 tonnes per year and Q(0) = 70,000 tonnes, we have:
3.0 * 10^5 = (Q(t) - 70,000) / t
Multiplying both sides by t:
3.0 * 10^5 * t = Q(t) - 70,000
Rearranging the equation:
Q(t) = 3.0 * 10^5 * t + 70,000
Now we need to substitute this expression for Q(t) into the original equation Q(t) = 10^4(t^2 + 15t + 70):
10^4(t^2 + 15t + 70) = 3.0 * 10^5 * t + 70,000
Expanding and rearranging the equation:
10^4t^2 + 1.5 * 10^5t + 7 * 10^5 = 3.0 * 10^5t + 70,000
Simplifying:
10^4t^2 - 1.5 * 10^5t + 6.3 * 10^5 = 0
This is a quadratic equation in t. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Substituting a = 10^4, b = -1.5 * 10^5, and c = 6.3 * 10^5 into the quadratic formula, we can find the values of t.
Note: Calculating the exact values using the quadratic formula may be quite complex due to the large numbers involved. It may be helpful to use approximation techniques or a calculator with high precision.
Once you find the values of t, you can check if they are within the range 0 to 10 (as given in the problem) to find the specific time when the rate of change reaches 3.0 * 10^5 tonnes per year.