A specialized coolant's temperature, θ\thetaθ degrees Celsius, in a high-performance engine is modeled by the equation
θ=225e−0.04t \theta = 225e^{-0.04t} θ=225e−0.04twhere ttt 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}dtdθ, in ∘C min−1^{\circ}\text{C min}^{-1}∘C min−1.
Select the correct answer from the options below:
dθdt=−9e−0.04t\frac{d\theta}{dt} = -9e^{-0.04t}dtdθ=−9e−0.04t
dθdt=9e−0.04t\frac{d\theta}{dt} = 9e^{-0.04t}dtdθ=9e−0.04t
dθdt=−5625e−0.04t\frac{d\theta}{dt} = -5625e^{-0.04t}dtdθ=−5625e−0.04t
dθdt=−0.04e−0.04t\frac{d\theta}{dt} = -0.04e^{-0.04t}dtdθ=−0.04e−0.04t
