a box with square base of length x and height y is changing shape such that \\( \\frac { d x } { d t } = 3…
a box with square base of length x and height y is changing shape such that \\( \\frac { d x } { d t } = 3 \\mathrm { cm } / \\mathrm { s }, \\frac { d y } { d t } = 7 \\mathrm { cm } / \\mathrm { s } \\). find the rate of change of volume when \\( x = 2 \\mathrm { cm } \\), \\( y = 4 \\mathrm { cm } \\).\nrecall: the volume v of a box with length \\( \\ell \\), width w, and height h is \\( v = \\ell w h \\).\n\\( 44 \\mathrm { cm } ^ { 3 } / \\mathrm { s } \\)\n\\( 89 \\mathrm { cm } ^ { 3 } / \\mathrm { s } \\)\n\\( 76 \\mathrm { cm } ^ { 3 } / \\mathrm { s } \\)\n\\( 31 \\mathrm { cm } ^ { 3 } / \\mathrm { s } \\)\n\\( 5 \\mathrm { cm } ^ { 3 } / \\mathrm { s } \\)
Answer
Explanation:
Step1: Write the volume formula
The volume (V) of the box with square base is (V = x^{2}y).
Step2: Differentiate (V) with respect to (t) using the product rule
The product rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = x^{2}) and (v = y). First, find (\frac{dV}{dt}): (\frac{dV}{dt}=\frac{d(x^{2}y)}{dt}=2xy\frac{dx}{dt}+x^{2}\frac{dy}{dt})
Step3: Substitute the given values
Given (x = 2\mathrm{cm}), (y = 4\mathrm{cm}), (\frac{dx}{dt}=3\mathrm{cm/s}), (\frac{dy}{dt}=7\mathrm{cm/s}) (\frac{dV}{dt}=2\times2\times4\times3+2^{2}\times7) (=48 + 28) (=76\mathrm{cm^{3}/s})
Answer:
(76\mathrm{cm^{3}/s}) (the third option)