1.9 Calculus

A specialized coolant's temperature, $θ\theta$ degrees Celsius, in a high-performance engine is modeled by the equation

\theta = 225e^{-0.04t}

where $t$ is the time in minutes since the engine was deactivated.

Determine an expression for the rate of change of the temperature, $dθdt\frac{d\theta}{dt}$ , in $min−1^{\circ}\text{C min}^{-1}$ .

Select the correct answer from the options below:

$dθdt=−9e−0.04t\frac{d\theta}{dt} = -9e^{-0.04t}$

$dθdt=9e−0.04t\frac{d\theta}{dt} = 9e^{-0.04t}$

$dθdt=−5625e−0.04t\frac{d\theta}{dt} = -5625e^{-0.04t}$

$dθdt=−0.04e−0.04t\frac{d\theta}{dt} = -0.04e^{-0.04t}$

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1.9 Calculus

1.9 Calculus Questions