Numerical methods: Applications to modelling
In a nutshell
The numerical methods used to find roots have direct applications in modelling real life scenarios. These models can be very useful in fields of science and engineering.
Example 1
The price of a motorcycle in £, x years after purchase is modelled using the following function:
f(x)=5000(0.75)x−300cos(x),x>0
(i) Use the model to find the value of the motorcycle after 5 years, to the nearest hundred.
(ii) Show that f(x) has a root between 11 and 12.
(iii) Find f′(x).
(iv) Taking 11.6 as an initial approximation, apply the Newton-Raphson method once to f(x) to get a second approximation for the time at which the motorcycle's value reaches 0. Give your answer to 3 decimal places.
(v) Comment on the validity of this model as an approximation for the motorcycle's value.
(i) Substitute x=5 into f(x).
f(5)=5000(0.75)5−300cos(5)=1101.424782
The motorcycle is worth £1100 after five years, rounded to the nearest hundred.
(ii) Substitute x=11 and x=12 into f(x).
f(11)=5000(0.75)11−300cos(11)=209.847...>0
f(12)=5000(0.75)12−300cos(12)=−94.774...<0
f(x) is continuous, and there is a change in sign in the interval [11,12]. f(x) must equal 0 within this interval, therefore:
f(x) has a root between 11 and 12.
(iii) Differentiate with respect to x.
f′(x)=300sin(x)+(5000)(0.75)x(ln0.75)
(iv) Substitute 11.6 into f(x) and f′(x).
f(11.6)f′(11.6)xn+1=5000(0.75)11.6−300cos(11.6)=7.210...=300sin(11.6)+(5000)(0.75)11.6(ln0.75)=520.357...=xn−f′(x)f(x)=11.6−520.357...7.210...=11.6241...=11.624 (3 s.f.)
(v) Use your quality of written communication to write a sentence about the real life implications of the model and its limitations.
Objects in real life cannot have a negative value. This model is not valid for motorcycles that are 12 years older or more.