What is Inverse Variation

In the given two variables if the value of one variable increases and another one decreases   that is if the values change in opposite manner then it’s said to be inverse variation.
For example if the speed increases the travelling time decreases and if speed decreases the travelling time increases

Inverse variations  occurs  when two variables such as XY are always equation to some constant K ie, one value  decreases and the other value increases

No of workers Days to complete the  work
        15               15
        10               18
         5               20

As the   number of worker decreases the days to complete the work increases.  Here inverse variations occur.

Inverse Variation Formula

Inverse-variation between two variables say  y and x is given by the formula    y  =  k/x
Here the variable K is known as the constant of proportionality.  Y varies inversely with x.
Y varies inversely with x, if there is a nonzero constant k such that xy = k or ,  y  = k/x, where k ≠ 0.
In inverse-variation its seen that , when the value of one  variable increases the value of the other value  decreases in proportion so that the product remains the same always.

Inverse Variation Word Problems

While solving inverse-variation word problems we use the formula ‘y = k/x   or any other variables relevant to the problem. Find the value of ‘k’ which is the constant of proportionality.
Plug in the value of k in the formula y=k/x .  Use the information from the given problem and solve for the variable.  An example problem is solved below.

If a car travels at the rate of 20 miles per hour it takes 1.5 hours to reach its destination.  If the same car travels at the rate of  50 miles per hour  what will be the time taken to reach its destination.

Use  the formula     t  =  k/s  where  t = 1.5 and s = 20 .  Plug in the values of t and s and find k.
         
                                   1.5  = k/20    (multiply by 20 both sides)
                                   K  = 1.5 x 20
                                   K =  30

Substitute the value of ‘k in the  equation   t = k/s
Here   k = 30  and  t  =  50
                                    t = 30/50

                                    t =  0.6 hours
The car takes 0.6 hours if it travels at the speed of 50 miles per hour.    Here it is found that as the speed of the car increases the time taken reduces.  So   there is inverse-variation here.

Dot product

There are two types of multiplication of two vectors. That is, as two types of products. One of them gives the result as a vector quantity and the other results as a scalar quantity. They are termed as cross product and dot product respectively.  The names are given so because in the first case a ‘x’ is used to indicate the multiplication and in the second case a ‘•’ indicates the same operation. Because of the fact that a dot product results in scalar quantity, the dot-product is also called as scalar dot product.

If P and Q are two vectors, then the dot-product two vectors is denoted as P • Q and its magnitude is given by lPl*lQlcos θ, where θ is the angle between the vectors P and Q. Unlike vector products, scalar products of vectors are commutative.

A typical example of a scalar product is the calculation of work done W by a force F while displacing the object by a displacement D. As per physics concept, W = lFl*lDlcos θ, where θ is the angle between the force and the direction of displacement. Although, F and D are vector quantities, the work done W is a scalar quantity. The work done by a force has no direction. .

From the definition, P • Q = lPl*lQlcos θ, let us analyze when the dot product is zero. It is obvious that it happens when P or Q or both are 0. But there is one more possibility. The scalar product becomes 0 when the angle between the vectors is 90o, even though P and Q may have finite values, since cos 90o = 0. In other words, the scalar product of two vectors acting at right angle is 0. We can illustrate this fact with a practical example. We all know that the weight W of an object placed on a smooth horizontal surface is a vector acting vertically. Suppose the object is moved horizontally by a displacement D, the vectors W and D act at right angle. Since the scalar product is 0, the work done W is 0. That is, no work is needed to be done in moving an object at a horizontal level subject to the assumption that there is no friction between the object and the horizontal surface. That is why we find that only a little amount of effort is needed to push even a heavy object horizontally. It is not zero as per theory, because the assumption of ‘no friction’ is practically impossible and the little effort only accounts for work done against the frictional force.