Assuming the petrol burnt (per hour) in driving a motor boat...

Assuming the petrol burnt (per hour) in driving a motor boat varies as the cube of its velocity, show that most economical speed when going against a current of $c \mathrm{~km}$ per hour is $\left(\frac{3 c}{2}\right) \mathrm{kms} /$ hour.

Key Concepts

Relative Velocity in Fluid Dynamics

In problems involving currents, the effective speed of a boat is not simply its speed in still water. Instead, when moving against a current, the actual speed relative to the ground is reduced by the speed of the current. This concept is crucial for analyzing scenarios where the movement of a medium (like water or air) influences the overall motion of an object.

Optimization Using Differential Calculus

Optimization in calculus involves finding the minimum or maximum values of a function by examining its derivatives. In the context of this problem, one must set up a function that relates the fuel consumption per unit distance to the boat's speed and then differentiate it with respect to that speed. Setting the derivative equal to zero allows us to determine the most economical speed that minimizes fuel consumption while going against the current.

Proportional Relationships

This concept involves understanding how one quantity varies as a power of another. In the problem, the fuel consumption (petrol burnt per hour) varies as the cube of the boat's velocity. Recognizing such power law relationships is fundamental in modeling real-life phenomena where changes in one variable produce nonlinear effects in another.

Transcript

00:01 Hello, we have a question which is based on the optimization of the cost and which is in turn application of the derivative only.

00:14 So it is given that the fuel cost per hour, that is fuel cost per hour is given as half of the cube of the speed measured in naught plus additional cost.

00:33 $216 per hour and so according to this model we have to optimize for for the cost total cost okay and we need to find the most economical speed when the distance travel days 500 miles 500 miles okay so let us suppose that it takes time t to travel 500 miles so nautical miles so nautical miles so so speed will be distance by time, that is 500 by t.

01:12 And since this cost is per hour, so we have to multiply both sides by t to get the cost for t hours.

01:22 So cost function, let us suppose cost c will become equal to, after multiplying with t, half v -cube -t plus 216 t.

01:37 Now let us plug in the value of v as this 500 by t so 1 by 2 into 500 by t whole cube into t plus 2116 t so this is 1 by 2 into 500 cube by t plus 216 t that is t squared so this will be 1 by 2 into 500 cube into t x2 t x2 x2 x2 x2 x2 x2 x2 2 plus 216 t this is the cost now we have to optimize this cost so for this we have to take we have to take the first derivative so this will be 1 by 2 into 500 cube into minus 2 t to the power minus 3 plus 2116 so these two will get cancelled out so we'll be writing 500 cube by t cube with negative sign of course 216 so we have to for critical points we have to equal this equal to 0 therefore 500 cube by t q equal to 216 so t cube will be equal to 500 cube by 216 so t cube will be equal to 500 cube by 216 which means t will be equal to 500 by 6 cube raised to the power 1 by 3.

03:13 Now these two will get cancelled out...

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