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
717 exam-style questions on Edexcel A Level Maths Differentiation, covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change. Each one has a worked solution and a mark scheme showing where the marks go.