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
425 exam-style questions on OCR A Level Maths 1.7 Differentiation, covering 1.7.1 Derivative as gradient of the tangent, 1.7.2 Gradient of the tangent at a point, 1.7.3 Sketching the gradient function, 1.7.4 Second derivatives, 1.7.5 Second derivative as rate of change of gradient, 1.7.6 Convex, concave and points of inflection (A-level only), 1.7.7 Differentiation from first principles for powers of x, 1.7.8 Differentiation from first principles for sin x and cos x (A-level only), 1.7.9 Differentiating x^n, 1.7.10 Differentiating e^(kx) and a^(kx) (A-level only), 1.7.11 Differentiating trigonometric functions (A-level only), 1.7.12 Derivative of ln x (A-level only), 1.7.13 Tangents and normals, 1.7.14 Stationary points, 1.7.15 Increasing and decreasing functions, 1.7.16 Points of inflection (A-level only), 1.7.17 Product and quotient rules (A-level only), 1.7.18 Chain rule (A-level only), 1.7.19 Parametric and implicit differentiation (A-level only), 1.7.20 Constructing differential equations (A-level only), and 1.7 Differentiation. Each one has a worked solution and a mark scheme showing where the marks go.