Arectangular 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 2.8 cm if d l divided by dt equals 1.9 times 10 superscript negative 5 baseline cm divided by degrees upper c and dw divided by dt equals 8.5 times 10 superscript negative 6 baseline cm divided by degrees c. round to one decimal place.

The variation rate is 5.42 10⁻⁵ cm²/ºC

Explanation:

When we have a thermal expansion problem we must have the relationship of the change in length as a function of the temperature, which are given in this problem, so we can write the expression for the area of a rectangle

      a = L W

They ask us to find the rate of variation of this area depending on the temperature, so we can derive this expression with respect to the temperature

    da / dT = d(LW) / dt

We use the derivative of a product since the two magnitudes change

    da / dT = W dL/dT + L dW/dT

The values they give us are

= 1.9 10⁻⁵ cm/ºC

= 8.5 10⁻⁶ cm/ºC

W = 1.6  cm

L= 2.8 cm

 Substituting the values and calculating

= 1.6 1.9 10⁻⁵ + 2.8 8.5 10⁻⁶

= 3.04 10⁻⁵ + 2.38 10⁻⁵

=  5.42 10⁻⁵ cm²/ºC

The variation rate is 5.42 10⁻⁵ cm²/ºC