Question

Energy consumption On the first day of the year \((t=0),\) a city uses electricity at a rate of \(2000 \mathrm{MW}\). That rate is projected to increase at a rate of \(1.3 \%\) per year. a. Based on these figures, find an exponential growth function for the power (rate of electricity use) for the city. b. Find the total energy (in MW-yr) used by the city over four full years beginning at \(t=0\) c. Find a function that gives the total energy used (in MW-yr) between \(t=0\) and any future time \(t>0\)

Short answer

Answer: The exponential growth function for the power of the city is \(P(t) = 2000 * (1 + 0.013)^t\). Over four years, the city uses approximately \(8448.83 \mathrm{MW\cdot yr}\) of energy.

Step-by-step solution

a. Find the exponential growth function

Let \(P(t)\) be the power in MW at time \(t\) in years. We are given that: 1. \(P(0) = 2000 \mathrm{MW}\) - The initial power at \(t=0\). 2. The growth rate is \(1.3\%\) per year. Therefore, the exponential growth function can be represented as: \(P(t) = P(0)* (1 + r)^t\), where \(r\) is the growth rate. We need to find \(P(t)\) given that \(r = 1.3\% = 0.013\).

a. Exponential growth function formula

Using the formula \(P(t) = P(0) * (1+r)^t\), we can find the exponential growth function for the power (rate of electricity use) for the city: \(P(t) = 2000 * (1 + 0.013)^t\)

b. Find the total energy used over four years

To find the total energy used over four years, we need to integrate the function \(P(t)\) with respect to time. The total energy used, \(E(t)\), can be represented as: \(E(t) = \int_0^t P(s)ds\) We need to find \(E(4)\) to calculate the total energy used over four years.

b. Calculate the integral of P(t)

First, we'll find the integral of \(P(t) = 2000 * (1+0.013)^t\): \(\int P(s)ds = \int 2000 * (1 + 0.013)^s ds\) To integrate this function, we'll make a substitution: \(u = 1 + 0.013 = 1.013\) \(du = 0.013 * ds\) \(ds = \frac{1}{0.013}du\) Now, substitute \(u\) and \(ds\): \(\int 2000 * u^s \frac{1}{0.013} du\) \(\frac{2000}{0.013} \int u^s du\) Now, we can find the integral: \(\frac{2000}{0.013} * \frac{u^{s+1}}{s+1} + C\) Now, replace \(u\): \(E(s) = \frac{153846.15 * (1.013)^{s+1}}{s+1} + C\)

b. Find the total energy used over four years

Now, we want to find the total energy used over four years: \(E(4) = \int_0^4 P(s)ds\) Substitute \(P(s)\): \(E(4) = \int_0^4 2000 * (1.013)^s ds\) Using the integral that we found in the previous step, we can find the total energy used. \(E(4) = \left[ \frac{153846.15 *(1.013)^{s+1}}{s+1} \right]_0^4\) \(E(4) = \frac{153846.15 * (1.013)^5}{5} - \frac{153846.15 * (1.013)^1}{1}\) Calculate the values: \(E(4) \approx 8448.83 \mathrm{MW\cdot yr}\) The total energy used by the city over four years is approximately \(8448.83 \mathrm{MW\cdot yr}\).

c. Find a function for the total energy used in t years

We've already found a function for the total energy used between \(t=0\) and any future time \(t>0\). The function is: \(E(t) = \frac{153846.15 * (1.013)^{t+1}}{t+1} - \frac{153846.15 * (1.013)^1}{1}\) This function can be used to calculate the total energy used between \(t=0\) and any future time \(t>0\).