The concentration of a chemical reactant in a solution, CCC mg/L, after ttt minutes is modeled by the equation:
log10C=1.84−0.072t \log_{10} C = 1.84 - 0.072t log10C=1.84−0.072tShow that this equation can be written in the form C=km−tC = km^{-t}C=km−t, where kkk and mmm are constants. Give the value of kkk to the nearest whole number and the value of mmm to 2 significant figures.
With reference to the equation in part (a), interpret the value of the constant kkk.
When the reaction temperature is increased, the concentration CCC after ttt minutes satisfies the equation:
C=560×1.15−t C = 560 \times 1.15^{-t} C=560×1.15−tUse calculus to find, to 2 significant figures, the value of dCdt\frac{dC}{dt}dtdC when t=4t = 4t=4.
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.