1. **Problem Statement:** We have a server response time modeled by the function $$T(n) = \alpha n^2 + \beta n + \gamma$$ where $n$ is the number of concurrent users. We need to find: (a) The first differential $dT$ when $n=100$. (b) The second differential $d^2T$. (c) Analyze how the differential changes with increasing $n$ and its implications for scalability. 2. **Formulas and Rules:** - The first differential $dT$ is the derivative of $T(n)$ with respect to $n$, multiplied by $dn$: $$dT = T'(n) dn$$ - The second differential $d^2T$ is the second derivative of $T(n)$ with respect to $n$, multiplied by $(dn)^2$: $$d^2T = T''(n) (dn)^2$$ - Here, $dn$ represents a small change in $n$. 3. **Step-by-step Solution:** **(a) Find $dT$ when $n=100$:** - Compute the first derivative: $$T'(n) = \frac{d}{dn}(\alpha n^2 + \beta n + \gamma) = 2\alpha n + \beta$$ - Evaluate at $n=100$: $$T'(100) = 2\alpha \times 100 + \beta = 200\alpha + \beta$$ - Therefore, the first differential is: $$dT = (200\alpha + \beta) dn$$ **(b) Find the second differential $d^2T$:** - Compute the second derivative: $$T''(n) = \frac{d}{dn}(2\alpha n + \beta) = 2\alpha$$ - The second differential is: $$d^2T = 2\alpha (dn)^2$$ **(c) Analysis of differential changes and scalability:** - The first derivative $T'(n) = 2\alpha n + \beta$ increases linearly with $n$ if $\alpha > 0$. - This means as the number of concurrent users increases, the rate of change of response time increases, indicating the server slows down more rapidly. - The second derivative $T''(n) = 2\alpha$ is constant, showing the acceleration of response time change is steady. - Implication: If $\alpha$ is positive and significant, the server's response time grows quadratically with users, which may limit scalability. - To improve scalability, reducing $\alpha$ (the quadratic coefficient) is crucial. **Final answers:** - $$dT = (200\alpha + \beta) dn$$ at $n=100$ - $$d^2T = 2\alpha (dn)^2$$ This completes the problem.