Answer to Question #302919 in Calculus for RAFAEL

Question #302919

When heating a 15 cm long square metal plate, its size increases by 0.05 cm.


Actions to be performed:


Answer: approximately how much did its area increase?


Determine how much its volume increases if it were a cubic plate.


If the plate is cooled and its side decreases by 0.05, how much did its area decrease?


If it is a cubic plate determined and it is cooled by decreasing its side by 0.05, determine how much its volume decreased?


1
Expert's answer
2022-03-02T16:51:46-0500

1. Δx=0.05 cm\Delta x=0.05\ cm


A(x)=x2A(x)=x^2

ΔA=A(x0+Δx)A(x0)=(x0+Δx)2(x0)2\Delta A=A(x_0+\Delta x)-A(x_0)=(x_0+\Delta x)^2-(x_0)^2

=(2x0+Δx)Δx=(2x_0+\Delta x)\Delta x

ΔA=(2(15cm)+0.05cm)(0.05cm)=1.5025cm2\Delta A=(2(15cm)+0.05cm)(0.05cm)=1.5025{cm}^2

The area increased by 1.5025 cm2.1.5025\ {cm}^2.



V(x)=x3V(x)=x^3

ΔV=V(x0+Δx)V(x0)=(x0+Δx)3(x0)3\Delta V=V(x_0+\Delta x)-V(x_0)=(x_0+\Delta x)^3-(x_0)^3

=((x0+Δx)2+(x0+Δx)x0+(x0)2)Δx=((x_0+\Delta x)^2+(x_0+\Delta x)x_0+(x_0)^2)\Delta x

ΔV=((15.05cm)2+15.05cm(15cm)+(15cm)2)(0.05cm)\Delta V=((15.05cm)^2+15.05cm(15cm)+(15cm)^2)(0.05cm)

=33.862625cm3=33.862625 cm^3

The volume increased by 33.862625cm3.33.862625 cm^3.


2. Δx=0.05 cm\Delta x=-0.05\ cm


A(x)=x2A(x)=x^2

ΔA=A(x0+Δx)A(x0)=(x0+Δx)2(x0)2\Delta A=A(x_0+\Delta x)-A(x_0)=(x_0+\Delta x)^2-(x_0)^2

=(2x0+Δx)Δx=(2x_0+\Delta x)\Delta x

ΔA=(2(15cm)0.05cm)(0.05cm)=1.4975cm2\Delta A=(2(15cm)-0.05cm)(-0.05cm)=-1.4975{cm}^2

The area decreased by 1.4975 cm2.1.4975\ {cm}^2.



V(x)=x3V(x)=x^3

ΔV=V(x0+Δx)V(x0)=(x0+Δx)3(x0)3\Delta V=V(x_0+\Delta x)-V(x_0)=(x_0+\Delta x)^3-(x_0)^3

=((x0+Δx)2+(x0+Δx)x0+(x0)2)Δx=((x_0+\Delta x)^2+(x_0+\Delta x)x_0+(x_0)^2)\Delta x

ΔV=((14.95cm)2+14.95cm(15cm)+(15cm)2)(0.05cm)\Delta V=((14.95cm)^2+14.95cm(15cm)+(15cm)^2)(-0.05cm)

=33.637625cm3=-33.637625 cm^3

The volume decreased by 33.637625cm3.33.637625 cm^3.



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