Let's solve **question 4** step-by-step: --- ### **4. An investment’s value, \( V(t) \), is modelled by the function \( V(t) = 2500(1.15)^t \), where \( t \) is the number of years after funds are invested.** #### **(a) To find the instantaneous rate of change in the value of the investment at \( t = 4 \), what intervals on either side of 4 would you choose? Why?** **Step 1:** - The instantaneous rate of change at \( t = 4 \) can be estimated by finding the average rate of change over a small interval around \( t = 4 \). - Typically, we choose intervals very close to 4, such as \([3.9, 4]\) and \([4, 4.1]\), or symmetrically, \([4 - h, 4 + h]\) for a small \( h \). **Sample Choice:** - Let's take \( t_1 = 3.9 \) and \( t_2 = 4.1 \). **Reason:** - The closer the points are to 4, the better the estimate of the instantaneous rate of change. Too large an interval will not accurately represent the slope at exactly \( t = 4 \). --- #### **(b) Use your intervals from part a) to find the instantaneous rate of change in the value of the investment at \( t = 4 \).** **Step 2:** - The average rate of change between \( t_1 \) and \( t_2 \) is: \[ \frac{V(t_2) - V(t_1)}{t_2 - t_1} \] **Step 3: Calculate \( V(3.9) \) and \( V(4.1) \):** - \( V(3.9) = 2500 \times (1.15)^{3.9} \) - \( V(4.1) = 2500 \times (1.15)^{4.1} \) Let's compute these values: #### \( (1.15)^{3.9} \) and \( (1.15)^{4.1} \): - \( (1.15)^{3.9} \approx e^{3.9 \cdot \ln(1.15)} \approx e^{3.9 \cdot 0.1398} \approx e^{0.5452} \approx 1.7245 \) - \( (1.15)^{4.1} \approx e^{4.1 \cdot \ln(1.15)} \approx e^{4.1 \cdot 0.1398} \approx e^{0.5732} \approx 1.7739 \) #### Now, calculate \( V(3.9) \) and \( V(4.1) \): - \( V(3.9) \approx 2500 \times 1.7245 = 4311.25 \) - \( V(4.1) \approx 2500 \times 1.7739 = 4434.75 \) **Step 4: Substitute into the average rate of change formula:** \[ \text{Average Rate of Change} = \frac{V(4.1) - V(3.9)}{4.1 - 3.9} \] \[ = \frac{4434.75 - 4311.25}{0.2} \] \[ = \frac{123.5}{0.2} \] \[ = 617.5 \] --- ### **Final Answer:** #### **4 (a):** Choose intervals very close to \( t = 4 \), such as \( [3.9, 4.1] \), because the closer the interval, the better the estimate of the instantaneous rate of change. #### **4 (b):** The instantaneous rate of change in the value of the investment at \( t = 4 \) is **approximately \$617.50 per year**. --- If you'd like the detailed solution to question 5 as well, let me know!

Let's solve **question 5** step-by-step: --- ### **5. The height \( y \), in centimeters, of a piston attached to a turning wheel at time \( t \), in seconds, is modeled by the equation:** \[ y = 2 \sin(120^\circ t) \] --- #### **(a) Examine the equation, and select a strategy for finding the instantaneous rate of change in the piston’s height at \( t = 12\, \text{s} \).** **Step 1: Understand the equation** - The height \( y(t) = 2 \sin(120^\circ t) \). - The sine function varies between -1 and 1, so \( y(t) \) varies between -2 and 2. - The argument of sine is \( 120^\circ t \), which indicates the function oscillates periodically. **Step 2: Identify the strategy** - To find the **instantaneous rate of change** at \( t = 12 \, \text{s} \), we need the **derivative** \( y'(t) \). - The derivative of \( y(t) = 2 \sin(120^\circ t) \) involves applying the chain rule: \[ y'(t) = 2 \times \cos(120^\circ t) \times \frac{d}{dt}(120^\circ t) \] - Since \( 120^\circ t \) is in degrees, convert degrees to radians for differentiation: \[ 120^\circ = \frac{120 \pi}{180} = \frac{2\pi}{3} \text{ radians} \] - Therefore, \[ y(t) = 2 \sin\left(\frac{2\pi}{3} t\right) \] - The derivative: \[ y'(t) = 2 \times \cos\left(\frac{2\pi}{3} t\right) \times \frac{d}{dt}\left(\frac{2\pi}{3} t\right) = 2 \times \cos\left(\frac{2\pi}{3} t\right) \times \frac{2\pi}{3} \] - Simplified: \[ y'(t) = \frac{4\pi}{3} \cos\left(\frac{2\pi}{3} t\right) \] **Summary of strategy:** - Find the derivative \( y'(t) \) using chain rule. - Evaluate \( y'(t) \) at \( t=12 \, \text{s} \). --- #### **(b) Use your strategy from part (a) to find the instantaneous rate of change at \( t=12\, \text{s} \).** **Step 1: Plug \( t=12 \) into the derivative:** \[ y'(12) = \frac{4\pi}{3} \cos\left(\frac{2\pi}{3} \times 12\right) \] **Step 2: Simplify inside the cosine:** \[ \frac{2\pi}{3} \times 12 = \frac{2\pi}{3} \times 12 = 2\pi \times 4 = 8\pi \] **Step 3: Evaluate \( \cos(8\pi) \):** - Since \( \cos(\theta) \) has period \( 2\pi \), and \( 8\pi \) is a multiple of \( 2\pi \): \[ \cos(8\pi) = \cos(0 + 4 \times 2\pi) = \cos(0) = 1 \] **Step 4: Calculate \( y'(12) \):** \[ y'(12) = \frac{4\pi}{3} \times 1 = \frac{4\pi}{3} \] **Numerical approximation:** \[ \frac{4\pi}{3} \approx \frac{4 \times 3.1416}{3} \approx \frac{12.5664}{3} \approx 4.1888 \] --- ### **Final answers:** **(a):** To find the instantaneous rate of change at \( t=12\, \text{s} \), differentiate \( y(t) \) to get \( y'(t) \), then evaluate at \( t=12 \). The strategy is to use the chain rule on the sine function with the argument \( \frac{2\pi}{3} t \). **(b):** The instantaneous rate of change of the piston’s height at \( t=12\, \text{s} \) is approximately **\( 4.19\, \text{cm/sec} \)**. --- If you'd like further clarification or additional problems, feel free to ask!