Respuesta :
Answer:
a.)[tex]\frac{∂V}{∂t} = 315 m^{3}s^{-1}[/tex]
b.)[tex]\frac{∂S}{∂t} = 124 m^{2}s^{-1}[/tex]
c.)[tex]\frac{∂D}{∂t} = -10ms^{-1}[/tex]
Explanation:
a.)
We know that
Volume = Length × Width × height
V = l × w × h
As volume is a dependent variable and its depends upon the length, width and height of the box. These three variable further dependent on time. So, to calculate the rate of change of volume, we need to take the partial derivative of volume with respect to time.
[tex]\frac{∂V}{∂t} = \frac{∂V}{∂l}\frac{∂l}{∂t} + \frac{∂V}{∂w}\frac{∂w}{∂t} + \frac{∂V}{∂h} \frac{∂h}{∂t}[/tex]
[tex]\frac{∂V}{∂t} = wh\frac{∂l}{∂t} + hl\frac{∂w}{∂t} + wl \frac{∂h}{∂t}[/tex]
[tex]\frac{∂V}{∂t} = (9)(9)(4) + (9)(1)4) + (9)(1)(-5)[/tex]
[tex]\frac{∂V}{∂t} = 315 m^{3}s^{-1}[/tex]
b.)
We know that
Surface area = 2(length × width + length × height + width × heigth)
S = 2( l × w + l × h + w × h )
Using the same technique as in part a
[tex]\frac{∂S}{∂t} = \frac{∂S}{∂l}\frac{∂l}{∂t} + \frac{∂S}{∂w}\frac{∂w}{∂t} + \frac{∂S}{∂h} \frac{∂h}{∂t}[/tex]
[tex]\frac{∂S}{∂t} = 2[(w\frac{∂l}{∂t} +l\frac{∂w}{∂t})+( l\frac{∂h}{∂t}+ h\frac{∂l}{∂t}) + (w \frac{∂h}{∂t}+ h \frac{∂w}{∂t})][/tex]
[tex]\frac{∂S}{∂t} = 2[(9.4+1.4)+( 1.-5+ 9.4) + (9.-5+ 9.4)][/tex]
[tex]\frac{∂S}{∂t} = 124 m^{2}s^{-1}[/tex]
c.)
We know that
Length of diagonal = (length)² × (height)² × (width)²
D = (l)²× (w)² × (h)²
Using the same technique as in part a
[tex]\frac{∂D}{∂t} = \frac{∂D}{∂l}\frac{∂l}{∂t} + \frac{∂D}{∂w}\frac{∂w}{∂t} + \frac{∂D}{∂h} \frac{∂h}{∂t}[/tex]
[tex]\frac{∂D}{∂t} = [2(l\frac{∂l}{∂t})+2( h\frac{∂h}{∂t})+ 2(w\frac{∂w}{∂t})][/tex]
[tex]\frac{∂D}{∂t} = [2(4)+2(-45)+ 2(36)][/tex]
[tex]\frac{∂D}{∂t} = -10ms^{-1}[/tex]
Minus sign indicate that length of diagonal decreases.
Part A. The rate change in the volume is 315 meter cube per second.
Part B. The rate change in the surface area is 22320 meters square per second.
Part C. The rate change in the length of the diagonal is 7.55 meters.
How do you calculate the volume, area and length of the diagonal?
Given that a box has length = 1 m, width = 9 m and height = 9 m. Also given that its length and width increase at a rate of 4 m/s and height is decreasing at a rate of 5 m/s.
Part A.
The Volume of the Box is calculated as,
Volume = Length [tex]\times[/tex] Width [tex]\times[/tex] Volume
Volume depends on the length, width and height of the box.
The change in the rate of volume can be calculated by the partial derivatives of variables of volume with respect to time.
[tex]\dfrac{dV}{dt} = \dfrac {dV}{dt}\dfrac {dl}{dt} + \dfrac {dV}{dt}\dfrac {dw}{dt} + \dfrac {dV}{dt}\dfrac {dh}{dt}[/tex]
[tex]\dfrac {dV}{dt} = wh \dfrac {dl}{dt} + lh \dfrac {dw}{dt} + lw \dfrac {dh}{dt}[/tex]
[tex]\dfrac {dV}{dt} = 9\times9\times 4 + 1\times 9\times 4 + 1\times 9\times (-5)[/tex]
[tex]\dfrac {dV}{dt} = 315 \;\rm m^3/s[/tex]
The rate change in the volume is 315 meter cube per second.
Part B.
The surface area is calculated as given below.
[tex]S = 2\times (l\times w + l\times h + w\times h)[/tex]
The change in the rate of surface area can be calculated by the partial derivatives of variables of the area with respect to time.
[tex]\dfrac {dS}{dt} = \dfrac {dS}{dt}\dfrac {dl}{dt} + \dfrac {dS}{dt}\dfrac {dw}{dt} + \dfrac {dS}{dt}\dfrac {dh}{dt}[/tex]
[tex]\dfrac {dS}{dt} = 2[ (w\dfrac {dl}{dt} + l\dfrac{dw}{dt}) \times( h\dfrac {dl}{dt} +l\dfrac{dh}{dt}) \times (w\dfrac {dh}{dt} + h\dfrac{dw}{dt})][/tex][tex]\dfrac {dS}{dt} = 2\times [(9\times 4 + 1 \times 4)\times (9\times 4 + 1 \times (-5)) \times (9\times(-5) + 9\times4)][/tex]
[tex]\dfrac {dS}{dt} = 22320 \;\rm m^2/s[/tex]
The rate change in the surface area is 22320 meter square per second.
Part C.
The diagonal of the cuboid box is
[tex]D = \sqrt{(l^2+w^2+h^2}[/tex]
The change in the rate of diagonal can be calculated by the partial derivatives of variables of diagonal with respect to the time.
[tex]\dfrac {dD}{dt} = \sqrt{(\dfrac {dl}{dt} )^2 + (\dfrac {dw}{dt} )^2 + (\dfrac {dh}{dt} )^2[/tex]
[tex]\dfrac {dD}{dt} = \sqrt{(4^2) + (4^2) + (-5^2)}[/tex][tex]\dfrac{dD}{dt}= 7.55 \;\rm m[/tex]
The rate change in the length of the diagonal is 7.55 meters.
For more information about the volume and area, follow the link given below.
https://brainly.com/question/46030.
