Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%
Solution and Explanation
The surface area of a cube (S) of side x is given by S = 6x2
\(\Rightarrow\)\(\frac{ds}{dx}\)=(\(\frac{ds}{dx}\))∇x
\(\Rightarrow\)(12x)∇x
\(\Rightarrow\)(12x)(0.01x) [ as1% of x is 0.01x]
\(\Rightarrow\)0.12x2
Hence, the approximate change in the surface area of the cube is 0.12x2 m2.
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Notes on Applications of Derivatives
Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
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