Answer to Question #302919 in Calculus for RAFAEL

Question

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?

Answer 1

1. $\Delta x=0.05\ cm$

$$A(x)=x^2$$

$$\Delta A=A(x_0+\Delta x)-A(x_0)=(x_0+\Delta x)^2-(x_0)^2$$

$$=(2x_0+\Delta x)\Delta x$$

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

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

$$V(x)=x^3$$

$$\Delta V=V(x_0+\Delta x)-V(x_0)=(x_0+\Delta x)^3-(x_0)^3$$

$$=((x_0+\Delta x)^2+(x_0+\Delta x)x_0+(x_0)^2)\Delta x$$

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

$$=33.862625 cm^3$$

The volume increased by $33.862625 cm^3.$

2. $\Delta x=-0.05\ cm$

$$A(x)=x^2$$

$$\Delta A=A(x_0+\Delta x)-A(x_0)=(x_0+\Delta x)^2-(x_0)^2$$

$$=(2x_0+\Delta x)\Delta x$$

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

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

$$V(x)=x^3$$

$$\Delta V=V(x_0+\Delta x)-V(x_0)=(x_0+\Delta x)^3-(x_0)^3$$

$$=((x_0+\Delta x)^2+(x_0+\Delta x)x_0+(x_0)^2)\Delta x$$

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

$$=-33.637625 cm^3$$

The volume decreased by $33.637625 cm^3.$