Solve the problem.
-The formula $\mathrm{E}=1000(100-\mathrm{T})+580(100-\mathrm{T})^{2}$ is used to approximate the elevation (in meters) above sea level at which water boils at a temperature of $\mathrm{T}$ (in degrees Celsius). Find the rate of change of $\mathrm{E}$ with respect to $\mathrm{T}$ for a temperature of $55^{\circ} \mathrm{C}$.
A) $-53,200 \mathrm{~m} /{ }^{\circ} \mathrm{C}$
B) $53,200 \mathrm{~m} /{ }^{\circ} \mathrm{C}$
C) $-52,200 \mathrm{~m} /{ }^{\circ} \mathrm{C}$
D) $-85,100 \mathrm{~m} /{ }^{\circ} \mathrm{C}$

Question

Quizplus · Accepted Answer

The rate of change of $E$ with respect to $T$ is the derivative $\frac{dE}{dT}$. Differentiating $E=1000(100-T)+580(100-T)^2$ with respect to $T$ gives $\frac{dE}{dT} = -1000 - 1160(100-T)$. Substituting $T=55$ gives $-1000 - 1160(45) = -53,200$.

Solve the Problem E=1000(100−T)+580(100−T)2\mathrm{E}=1000(100-\mathrm{T})+580(100-\mathrm{T})^{2}E=1000(100−T)+580(100−T)2 Is Used to Approximate the Elevation (In Meters) Above Sea

Solve the Problem $\mathrm{E}=1000(100-\mathrm{T})+580(100-\mathrm{T})^{2}$ Is Used to Approximate the Elevation (In Meters) Above Sea