Solve the problem.
-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 $1.1 \mathrm {~cm}$ and the length is $2.4 \mathrm {~cm}$ if $\mathrm { dl } / \mathrm { dt } = 1.9 \times 10 ^ { - 5 } \mathrm {~cm} / { } ^ { \circ } \mathrm { C }$ and $\mathrm { dw } / \mathrm { dt } = 8.5 \times 10 ^ { - 6 } \mathrm {~cm} / { } ^ { \circ } \mathrm { C }$.
A) $1.6 \times 10 ^ { - 5 } \mathrm {~cm} ^ { 2 } / { } ^ { \circ } \mathrm { C }$
B) $2.6 \times 10 ^ { - 5 } \mathrm {~cm} ^ { 2 } / { } ^ { \circ } \mathrm { C }$
C) $2.1 \times 10 ^ { - 5 } \mathrm {~cm} ^ { 2 } / { } ^ { \circ } \mathrm { C }$
D) $4.1 \times 10 ^ { - 5 } \mathrm {~cm} ^ { 2 } / { } ^ { \circ } \mathrm { C }$

Question

Quizplus · Accepted Answer

The rate of change of area with respect to temperature, $\frac{dA}{dT}$, can be found using the product rule for differentiation: $\frac{dA}{dT} = l\frac{dw}{dT} + w\frac{dl}{dT}$. Substituting the given values, $l = 2.4 \, \text{cm}$, $w = 1.1 \, \text{cm}$, $\frac{dl}{dT} = 1.9 \times 10^{-5} \, \text{cm/}^\circ\text{C}$, and $\frac{dw}{dT} = 8.5 \times 10^{-6} \, \text{cm/}^\circ\text{C}$, we get $\frac{dA}{dT} = (2.4)(8.5 \times 10^{-6}) + (1.1)(1.9 \times 10^{-5}) = 4.1 \times 10^{-5} \, \text{cm}^2/^\circ\text{C}$.

Solve the Problem TTT When the Width Is 1.1 cm1.1 \mathrm {~cm}1.1 cm And the Length Is

Solve the Problem $T$ When the Width Is $1.1 \mathrm {~cm}$ And the Length Is