The length of a rectangle is increasing at the rate of 2 feet per second, while the width is increasing at the rate of 1 foot per second. When the length is 4 feet and the width is 3 feet, how fast is the length of the diagonal of the rectangle increasing?
A) $1 \mathrm { ft } / \mathrm { s }$

B) $1.5 \mathrm { ft } / \mathrm { s }$
C) $2.5 \mathrm { ft } / \mathrm { s }$
D) $1.2 \mathrm { ft } / \mathrm { s }$
E)
$2 \mathrm { ft } / \mathrm { s }$
F)$2.3 \mathrm { ft } / \mathrm { s }$

G) $2.2 \mathrm { ft } / \mathrm { s }$

H) $3 \mathrm { ft } / \mathrm { s }$

Question

The length of a rectangle is increasing at the rate of 2 feet per second, while the width is increasing at the rate of 1 foot per second. When the length is 4 feet and the width is 3 feet, how fast is the length of the diagonal of the rectangle increasing?
A) $1 \mathrm { ft } / \mathrm { s }$

B) $1.5 \mathrm { ft } / \mathrm { s }$ 
C) $2.5 \mathrm { ft } / \mathrm { s }$ 
D) $1.2 \mathrm { ft } / \mathrm { s }$ 
E)
$2 \mathrm { ft } / \mathrm { s }$
F)$2.3 \mathrm { ft } / \mathrm { s }$

G) $2.2 \mathrm { ft } / \mathrm { s }$

H) $3 \mathrm { ft } / \mathrm { s }$

Quizplus · Accepted Answer

The length of the diagonal of a rectangle is given by the formula $d = \sqrt{l^2 + w^2}$. Taking the derivative of both sides with respect to time, we get $\frac{dd}{dt} = \frac{1}{2\sqrt{l^2 + w^2}}(2l\frac{dl}{dt} + 2w\frac{dw}{dt})$. Plugging in the given values, we get $\frac{dd}{dt} = \frac{1}{2\sqrt{4^2 + 3^2}}(2(4)(2) + 2(3)(1)) = 2.2 \mathrm { ft } / \mathrm { s }$.

The Length of a Rectangle Is Increasing at the Rate 1ft/s1 \mathrm { ft } / \mathrm { s }1ft/s

The Length of a Rectangle Is Increasing at the Rate $1 \mathrm { ft } / \mathrm { s }$