Answer:
5.832 g/h
Explanation:
We can find the rate of mass relative to time by taking the derivative of M
[tex]M' = \frac{dM}{dt} = \frac{dM}{dx}\frac{dx}{dt}[/tex]
[tex]M' = \frac{d(x^3 + 0.1x^4)}{dx}\frac{dx}{dt}[/tex]
[tex]M' = (3x^2 + 0.4x^3)x'[/tex]
We can plug in the current length x = 6 cm and current length rate 0.03 cm/h
[tex]M' = (3*6^2 + 0.4*6^3)0.03 = 5.832 g/h[/tex]
Answer:
The rate of mass of the cube increasing by [tex]5.832\rm g/hr[/tex]
Explanation:
Given information:
Mass of a cube [tex]M=x^3+0.1x^4[/tex]
[tex]\frac{dx}{dt}=0.03\rm cm/hr[/tex]
[tex]x[/tex] is the length in [tex]\rm cm[/tex]
By taken derivative of mass with respect to time [tex]\frac{dM}{dt}[/tex]
We can write it as ,
[tex]M'=\frac{dM}{dt}=\frac{dM}{dx}\frac{dx}{dt}[/tex]
[tex]M'=\frac{d(x^3+0.1x^4)}{dx} \frac{dx}{dt}[/tex]
[tex]M'=(3x^2+0.4x^3)\frac{dx}{dt}[/tex]
On substitution[tex]x=\rm 6cm[/tex]
[tex]M'=(3x^2+0.4x^3)\frac{dx}{dt}=(3(6)^2+0.4(6)^3)\times 0.03\rm cm/hr=5.832g/hr[/tex]
Hence, the rate of mass of the cube increasing by [tex]5.832\rm g/hr[/tex]
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