1. **State the problem:** We have a piecewise function for the emission factor $f(t)$ during peak hours: $$ f(t) = \begin{cases} 0.2 + k t e^{-0.0005 t^2}, & 0 \leq t < 30 \\ 0.5 + 0.05 \cos\left(\frac{\pi t}{60}\right) \cos\left(\frac{\pi t}{30}\right), & 30 \leq t < 90 \\ \frac{0.3}{1 + (t - 90)^2} + 0.2, & 90 \leq t \leq 120 \end{cases} $$ We need to find: - (a) The constant $k$ such that $f(t)$ is continuous on $[0,120]$. - (b) The average emission factor over $[0,120]$ using the found $k$. --- 2. **Find $k$ by enforcing continuity at $t=30$:** Continuity means: $$ \lim_{t \to 30^-} f(t) = f(30) = \lim_{t \to 30^+} f(t) $$ - Left limit at $t=30$ (from first piece): $$ f(30^-) = 0.2 + k \cdot 30 \cdot e^{-0.0005 \cdot 30^2} = 0.2 + 30k e^{-0.0005 \cdot 900} $$ Calculate the exponent: $$ -0.0005 \times 900 = -0.45 $$ So: $$ f(30^-) = 0.2 + 30k e^{-0.45} $$ - Right limit at $t=30$ (from second piece): $$ f(30^+) = 0.5 + 0.05 \cos\left(\frac{\pi \times 30}{60}\right) \cos\left(\frac{\pi \times 30}{30}\right) = 0.5 + 0.05 \cos\left(\frac{\pi}{2}\right) \cos(\pi) $$ Recall: $$ \cos\left(\frac{\pi}{2}\right) = 0, \quad \cos(\pi) = -1 $$ Therefore: $$ f(30^+) = 0.5 + 0.05 \times 0 \times (-1) = 0.5 $$ - Set equal for continuity: $$ 0.2 + 30k e^{-0.45} = 0.5 $$ - Solve for $k$: $$ 30k e^{-0.45} = 0.5 - 0.2 = 0.3 $$ $$ k = \frac{0.3}{30 e^{-0.45}} = \frac{0.3}{30} e^{0.45} = 0.01 e^{0.45} $$ Calculate $e^{0.45}$: $$ e^{0.45} \approx 1.5683 $$ So: $$ k \approx 0.01 \times 1.5683 = 0.0157 $$ Rounded to three significant figures: $$ k = 0.0157 $$ --- 3. **Calculate the average emission factor over $[0,120]$:** The average value is: $$ \overline{f} = \frac{1}{120 - 0} \int_0^{120} f(t) dt = \frac{1}{120} \left( \int_0^{30} f(t) dt + \int_{30}^{90} f(t) dt + \int_{90}^{120} f(t) dt \right) $$ - First integral: $$ \int_0^{30} \left(0.2 + 0.0157 t e^{-0.0005 t^2}\right) dt = \int_0^{30} 0.2 dt + 0.0157 \int_0^{30} t e^{-0.0005 t^2} dt $$ Calculate each part: $$ \int_0^{30} 0.2 dt = 0.2 \times 30 = 6 $$ For the second integral, use substitution: Let $u = -0.0005 t^2$, then $du = -0.001 t dt$ or $t dt = -\frac{du}{0.001}$. Change limits: - When $t=0$, $u=0$. - When $t=30$, $u = -0.0005 \times 900 = -0.45$. So: $$ \int_0^{30} t e^{-0.0005 t^2} dt = \int_0^{-0.45} e^{u} \left(-\frac{1}{0.001}\right) du = -1000 \int_0^{-0.45} e^{u} du $$ Flip limits to get rid of minus: $$ = 1000 \int_{-0.45}^0 e^{u} du = 1000 \left[e^{u}\right]_{-0.45}^0 = 1000 (1 - e^{-0.45}) $$ Calculate $e^{-0.45} \approx 0.6376$: $$ 1000 (1 - 0.6376) = 1000 \times 0.3624 = 362.4 $$ Multiply by 0.0157: $$ 0.0157 \times 362.4 = 5.69 $$ So the first integral is: $$ 6 + 5.69 = 11.69 $$ - Second integral: $$ \int_{30}^{90} \left(0.5 + 0.05 \cos\left(\frac{\pi t}{60}\right) \cos\left(\frac{\pi t}{30}\right)\right) dt = \int_{30}^{90} 0.5 dt + 0.05 \int_{30}^{90} \cos\left(\frac{\pi t}{60}\right) \cos\left(\frac{\pi t}{30}\right) dt $$ Calculate first part: $$ 0.5 \times (90 - 30) = 0.5 \times 60 = 30 $$ For the second part, use product-to-sum identity: $$ \cos A \cos B = \frac{1}{2} [\cos(A-B) + \cos(A+B)] $$ Let: $$ A = \frac{\pi t}{60}, \quad B = \frac{\pi t}{30} $$ Then: $$ \cos\left(\frac{\pi t}{60}\right) \cos\left(\frac{\pi t}{30}\right) = \frac{1}{2} \left[ \cos\left(\frac{\pi t}{60} - \frac{\pi t}{30}\right) + \cos\left(\frac{\pi t}{60} + \frac{\pi t}{30}\right) \right] $$ Simplify inside cosines: $$ \frac{\pi t}{60} - \frac{\pi t}{30} = \frac{\pi t}{60} - \frac{2\pi t}{60} = -\frac{\pi t}{60} $$ $$ \frac{\pi t}{60} + \frac{\pi t}{30} = \frac{\pi t}{60} + \frac{2\pi t}{60} = \frac{3\pi t}{60} = \frac{\pi t}{20} $$ So: $$ \cos\left(\frac{\pi t}{60}\right) \cos\left(\frac{\pi t}{30}\right) = \frac{1}{2} \left[ \cos\left(-\frac{\pi t}{60}\right) + \cos\left(\frac{\pi t}{20}\right) \right] $$ Since $\cos$ is even: $$ = \frac{1}{2} \left[ \cos\left(\frac{\pi t}{60}\right) + \cos\left(\frac{\pi t}{20}\right) \right] $$ Therefore: $$ 0.05 \int_{30}^{90} \cos\left(\frac{\pi t}{60}\right) \cos\left(\frac{\pi t}{30}\right) dt = 0.05 \times \frac{1}{2} \int_{30}^{90} \left[ \cos\left(\frac{\pi t}{60}\right) + \cos\left(\frac{\pi t}{20}\right) \right] dt $$ $$ = 0.025 \left( \int_{30}^{90} \cos\left(\frac{\pi t}{60}\right) dt + \int_{30}^{90} \cos\left(\frac{\pi t}{20}\right) dt \right) $$ Calculate each integral: - For $\int \cos(\alpha t) dt = \frac{1}{\alpha} \sin(\alpha t)$. First integral: $$ \int_{30}^{90} \cos\left(\frac{\pi t}{60}\right) dt = \left[ \frac{60}{\pi} \sin\left(\frac{\pi t}{60}\right) \right]_{30}^{90} = \frac{60}{\pi} \left( \sin\left(\frac{3\pi}{2}\right) - \sin\left(\frac{\pi}{2}\right) \right) $$ Recall: $$ \sin\left(\frac{3\pi}{2}\right) = -1, \quad \sin\left(\frac{\pi}{2}\right) = 1 $$ So: $$ = \frac{60}{\pi} (-1 - 1) = \frac{60}{\pi} (-2) = -\frac{120}{\pi} $$ Second integral: $$ \int_{30}^{90} \cos\left(\frac{\pi t}{20}\right) dt = \left[ \frac{20}{\pi} \sin\left(\frac{\pi t}{20}\right) \right]_{30}^{90} = \frac{20}{\pi} \left( \sin\left(\frac{9\pi}{2}\right) - \sin\left(\frac{3\pi}{2}\right) \right) $$ Recall: $$ \sin\left(\frac{9\pi}{2}\right) = \sin\left(4\pi + \frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 $$ $$ \sin\left(\frac{3\pi}{2}\right) = -1 $$ So: $$ = \frac{20}{\pi} (1 - (-1)) = \frac{20}{\pi} \times 2 = \frac{40}{\pi} $$ Sum of integrals: $$ -\frac{120}{\pi} + \frac{40}{\pi} = -\frac{80}{\pi} $$ Multiply by 0.025: $$ 0.025 \times \left(-\frac{80}{\pi}\right) = -\frac{2}{\pi} \approx -0.6366 $$ So the second integral is: $$ 30 - 0.6366 = 29.3634 $$ - Third integral: $$ \int_{90}^{120} \left( \frac{0.3}{1 + (t - 90)^2} + 0.2 \right) dt = \int_{90}^{120} \frac{0.3}{1 + (t - 90)^2} dt + \int_{90}^{120} 0.2 dt $$ Calculate the constant part: $$ 0.2 \times (120 - 90) = 0.2 \times 30 = 6 $$ For the other integral, substitute $x = t - 90$, so when $t=90$, $x=0$; when $t=120$, $x=30$: $$ \int_0^{30} \frac{0.3}{1 + x^2} dx = 0.3 \int_0^{30} \frac{1}{1 + x^2} dx = 0.3 \left[ \arctan x \right]_0^{30} = 0.3 (\arctan 30 - 0) $$ Calculate $\arctan 30$ in radians (approx): $$ \arctan 30 \approx 1.537 $$ So: $$ 0.3 \times 1.537 = 0.461 $$ Sum of third integral: $$ 6 + 0.461 = 6.461 $$ --- 4. **Sum all integrals and find average:** $$ \int_0^{120} f(t) dt = 11.69 + 29.3634 + 6.461 = 47.5144 $$ Average emission factor: $$ \overline{f} = \frac{47.5144}{120} = 0.396 $$ Rounded to three significant figures: $$ \boxed{0.396} $$ --- **Final answers:** - (a) $k = 0.0157$ - (b) Average emission factor $= 0.396$