A rectangular steel plate expands as it is heated. Find the rate of change of area with respect to temperature T when the width is 1.6 cm and the length is 2.6 cm if d l divided by dt equals 1.1 times 10 Superscript negative 5 Baseline cm divided by degrees Upper C and dw divided by dt equals 8.9 times 10 Superscript negative 6 Baseline cm divided by degrees C. Round to one decimal place.

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Answer:

The variation rate is 4.07 × 10⁻⁵ cm²/ºC

Step-by-step explanation:

The relationship of the change in length as a function of the temperature, which are given in this problem can be written by the expression for the area of ​​a rectangle

a = L × W

Differentiating both sides,

[tex]\frac{da}{dT}[/tex] = [tex]\frac{d(LW)}{dT}[/tex]

[tex]\frac{da}{dT}[/tex] = W [tex]\frac{dL}{dT}[/tex] + L [tex]\frac{dW}{dT}[/tex]

The values ​​they give us are

[tex]\frac{dL}{dT}[/tex] = 1.1 × 10⁻⁵ cm/ºC

[tex]\frac{dW}{dT}[/tex] = 8.9 × 10⁻⁶ cm/ºC

W = 1.6  cm

L= 2.6 cm

Substituting the values ​​and calculating

[tex]\frac{da}{dT}[/tex] = (1.6 × 1.1 × 10⁻⁵) + (2.6 × 8.9 × 10⁻⁶)

[tex]\frac{da}{dT}[/tex] = (1.76 × 10⁻⁵) + (2.31 × 10⁻⁵)

[tex]\frac{da}{dT}[/tex] =  4.07 × 10⁻⁵ cm²/ºC

The variation rate is 4.07 × 10⁻⁵ cm²/ºC

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