At what rate is the diagonal of a cube increasing if its edges are 2 inches and are increasing at a rate 3 inches per minute?
A) $\sqrt { 3 } \mathrm { in } / \mathrm { min }$

B) $6 \sqrt { 3 } \text { in } / \min$
C) $2 \sqrt { 3 } \mathrm { in } / \mathrm { min }$
D) $72 \text { in } / \mathrm { min }$
E) $3 \sqrt { 3 } \text { in } / \min$
F) $9 \mathrm { in } / \mathrm { min }$

G) $36 \text { in } / \mathrm { min }$

H) $24 \text { in } / \mathrm { min }$

Question

At what rate is the diagonal of a cube increasing if its edges are 2 inches and are increasing at a rate 3 inches per minute?
A) $\sqrt { 3 } \mathrm { in } / \mathrm { min }$

B) $6 \sqrt { 3 } \text { in } / \min$ 
C) $2 \sqrt { 3 } \mathrm { in } / \mathrm { min }$ 
D) $72 \text { in } / \mathrm { min }$ 
E) $3 \sqrt { 3 } \text { in } / \min$ 
F) $9 \mathrm { in } / \mathrm { min }$

G) $36 \text { in } / \mathrm { min }$

H) $24 \text { in } / \mathrm { min }$

Quizplus · Accepted Answer

The diagonal of a cube with edge length $a$ is given by $d = a\sqrt{3}$. The rate of change of the diagonal with respect to time, $\frac{dd}{dt}$, can be found using the derivative of $d$ with respect to $a$ times the rate of change of $a$ with respect to time, $\frac{da}{dt}$. Given $\frac{da}{dt} = 3 \text{ in/min}$, we find $\frac{dd}{dt} = \sqrt{3} \times 3 = 3\sqrt{3} \text{ in/min}$.

At What Rate Is the Diagonal of a Cube Increasing 3in/min\sqrt { 3 } \mathrm { in } / \mathrm { min }3​in/min

At What Rate Is the Diagonal of a Cube Increasing $\sqrt { 3 } \mathrm { in } / \mathrm { min }$