sodium chlorate crystals are easy to grow in the shape of cubes by allowing a solution of water and sodium…

sodium chlorate crystals are easy to grow in the shape of cubes by allowing a solution of water and sodium chlorate to evaporate slo\nif ( v ) is the volume of such a cube with side length ( x ), find ( \frac{dv}{dx} ) (in ( mathrm{mm}^{3}/mathrm{mm} )) when ( x = 5 mathrm{~mm} ).\n( v(5)=quad mathrm{mm}^{3} / mathrm{mm} )\nexplain the meaning of ( v^{prime}(5) ) in the context of this problem.\n( v(5) ) represents the volume as the side length reaches ( 5 mathrm{~mm} ).\n( v(5) ) represents the rate at which the side length is increasing with respect to the volume as ( x ) reaches ( 5 mathrm{~mm} ).\n( v(5) ) represents the rate at which the volume is increasing as ( x ) reaches ( 15 mathrm{~mm} ).\n( v(5) ) represents the rate at which the volume is increasing with respect to the side length as ( x ) reaches ( 5 mathrm{~mm} ).\n( v(5) ) represents the rate at which the volume is increasing with respect to the side length as ( v ) reaches ( 15 mathrm{~mm}^{3} ).

sodium chlorate crystals are easy to grow in the shape of cubes by allowing a solution of water and sodium chlorate to evaporate slo\nif ( v ) is the volume of such a cube with side length ( x ), find ( \frac{dv}{dx} ) (in ( mathrm{mm}^{3}/mathrm{mm} )) when ( x = 5 mathrm{~mm} ).\n( v(5)=quad mathrm{mm}^{3} / mathrm{mm} )\nexplain the meaning of ( v^{prime}(5) ) in the context of this problem.\n( v(5) ) represents the volume as the side length reaches ( 5 mathrm{~mm} ).\n( v(5) ) represents the rate at which the side length is increasing with respect to the volume as ( x ) reaches ( 5 mathrm{~mm} ).\n( v(5) ) represents the rate at which the volume is increasing as ( x ) reaches ( 15 mathrm{~mm} ).\n( v(5) ) represents the rate at which the volume is increasing with respect to the side length as ( x ) reaches ( 5 mathrm{~mm} ).\n( v(5) ) represents the rate at which the volume is increasing with respect to the side length as ( v ) reaches ( 15 mathrm{~mm}^{3} ).

Answer

Explanation:

Step1: Find the volume formula of a cube

The volume formula of a cube is (V = x^{3}).

Step2: Differentiate the volume function with respect to (x)

Using the power rule ((x^{n})^\prime=nx^{n - 1}), for (V(x)=x^{3}), we have (\frac{dV}{dx}=V^\prime(x)=3x^{2}).

Step3: Substitute (x = 5) into the derivative

When (x = 5), (V^\prime(5)=3\times5^{2}=3\times25 = 75).

For the meaning of (V^\prime(5)): The derivative (\frac{dV}{dx}) represents the rate of change of the volume (V) with respect to the side - length (x). So (V^\prime(5)) represents the rate at which the volume is increasing with respect to the side length as (x) reaches (5) mm.

Answer:

(V(5)=75) (mm^{3}/mm) For the multiple - choice part: The correct option is (V^\prime(5)) represents the rate at which the volume is increasing with respect to the side length as (x) reaches (5) mm.