related rates in this section we will discussed

Question

Related Rates : In this section we will discussed for application of implicit differentiation. 

For these related rates problems usually it's best to just see some problems and see how they work.

Example: Air is pumped in a spherical balloon at a rate of 5 cm³/min. Find out the rate at which the radius of the balloon is raising while the diameter of the balloon is 20 cm.

Solution : The first thing that we'll have to do here is to recognize what information that we've been provided and what we desire to find. Previous to we do that let's notice that both of the volume of the balloon & the radius of the balloon will differ with time and thus are really functions of time, i.e. V (t ) and r (t ) .

We know that air is being pumped in the balloon at a rate of 5 cm³/min. It is the rate on which the volume is raising.  Recall that rates of change are derivatives and thus we know that,

V ′ (t ) = 5

We desire to find out the rate at which the radius is changing.  Again, rates are derivatives and thus it looks like we desire to determine,

                      r′ (t ) = ?            when           r (t ) = d /2= 10 cm

Note that we required converting the diameter to a radius.

Now that we've recognized what we have been given and what we desire to determine we have to relate these two quantities to each of other.  In this case we can relate the volume and the radius along with the formula for the volume of any sphere.

                                                        V (t ) = 4/3 ∏ [r (t )]³

As in the earlier section while we looked at implicit differentiation, typically we will not use the  (t ) part of things in the formulas, however since this is the first time through one of these we will do that to remind ourselves that they are actually functions of t.

Now we don't in fact want a relationship among the volume & the radius.  What we actually desire is a relationship among their derivatives.  We can accomplish this by differentiating both of the sides with respect to t.  In other terms, we will have to do implicit differentiation on the above formula. By doing this we get,

                                                             V ′ = 4 ∏ r ² r′

Note as well that at this point we went ahead and dropped the (t ) from each terms.  Now all that we have to do is plug in what we know and solve out for what we desire to find.

5 = 4 ∏ (10² ) r′           ⇒ r′ = 1 /80 ∏ cm/min

We can get the units of the derivative through recalling that,

 r′ = dr /dt