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Question: Using differentials, find the approximate value of the following: \(\sqrt {0.0082} \) A) \(0.901...

Using differentials, find the approximate value of the following:
0.0082\sqrt {0.0082}
A) 0.9010.901
B) 0.00090.0009
C) 0.09010.0901
D) 0.00190.0019

Explanation

Solution

The most important step in the calculation is to identify the correct function which we can use to find the given value or approximate value of the required function. Then we need to establish the approximate value of any number with respect to the given number and then we will use the definition of the differential to find the required value.

Complete step-by-step answer:
It is given that we have to find the value of 0.0082\sqrt {0.0082} .
It can be observed that the function is expressed as a square root so we will select the main function accordingly.
We will select the function yy as follows:
y=xy = \sqrt x
On differentiating with respect to xx we get the following:
dydx=12x\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt x }}
Now using the definition of the differential we will right the value of differential of yy as follows:
Δy=dydxΔx\Delta y = \dfrac{{dy}}{{dx}}\Delta x
Now the differential with respect to the variable xx that is Δx\Delta x the infinitesimal change in variable xx .
We will have to find the nearest value to the number 0.00820.0082 which square root we already know.
Therefore, we will consider x=0.0081x = 0.0081 that will imply that the small change is just 0.00010.0001 .
This will imply that Δx=0.0001\Delta x = 0.0001.
Now in this case Δy=0.0082\Delta y = \sqrt {0.0082} .
Now we express the given function as follows:
Δy=(dydx)0.0081(0.0001)\Delta y = {\left( {\dfrac{{dy}}{{dx}}} \right)_{0.0081}}\left( {0.0001} \right)
On simplifying the equation, we get the following:
Δy=(120.0081)(0.0001)\Delta y = \left( {\dfrac{1}{{2\sqrt {0.0081} }}} \right)\left( {0.0001} \right)
Therefore,
Δy=0.0901\Delta y = 0.0901
Thus, the answer to the given problem is 0.09010.0901 .

Hence, the correct option is A.

Note: The differentials always give an approximate value so we have used the approximate value of the dependent variable as well. We differentiate the given function and then substitute the known values to simplify the given values and reach the final answer.