Respuesta :
Answer:
A) 14 cm² per sec
B) 0 cm per sec
C) -28 cm per sec
Step-by-step explanation:
We know that,
If l = length of a rectangle and w = width of the rectangle
A) Area of a rectangle,
[tex]A=l\times w[/tex]
Differentiating with respect to t ( time )
[tex]\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}[/tex]
We have,
[tex]l=12\text{ cm}, w=5\text{ cm}, \frac{dw}{dt}=2\text{ cm per sec}, \frac{dl}{dt}=-2\text{ cm per sec}[/tex]
[tex]\frac{dA}{dt}=12\times 2+5\times -2[/tex]
[tex]\frac{dA}{dt}=24-10[/tex]
[tex]\frac{dA}{dt}=14\text{ square cm per sec}[/tex]
B) Perimeter of the rectangle,
[tex]P=2(l+w)[/tex]
Differentiating with respect to t ( time ),
[tex]\frac{dP}{dt}=2(\frac{dl}{dt}+\frac{dw}{dt})=2(-2+2)=0[/tex]
C) Length of the diagonal,
[tex]D=\sqrt{l^2+w^2}[/tex]
[tex]D^2 = l^2 + w^2[/tex]
Differentiating with respect to t ( time ),
[tex]2D\frac{dD}{dt}=2l\frac{dl}{dt}+2w\frac{dw}{dt}[/tex]
Since, if l = 12 cm, w = 5 cm,
[tex]D=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ cm}[/tex]
[tex]\implies 2\times 13 \frac{dD}{dt}=2(12)(-2)+2(5)(2)[/tex]
[tex]26\frac{dD}{dt}=-48+20=-28\text{ cm per sec}[/tex]
Using implicit differentiation, it is found that:
a) The rate of change of the area of the rectangle is: 14 cm²/sec.
b) The rate of change of the perimeter of the rectangle is: 0 cm/sec.
c) The rate of change of the lengths of the diagonal of the rectangle is: [tex]\mathbf{-\frac{14}{13}}[/tex] cm/sec.
Item a:
The area of a rectangle of length l and width w is given by:
[tex]A = lw[/tex]
Applying implicit differentiation, the rate of change is of:
[tex]\frac{dA}{dt} = w\frac{dl}{dt} + l\frac{dw}{dt}[/tex]
For this problem, the values are:
[tex]w = 5, \frac{dl}{dt} = -2, l = 12, \frac{dw}{dt} = 2[/tex]
Then:
[tex]\frac{dA}{dt} = 5(-2) + 12(2)[/tex]
[tex]\frac{dA}{dt} = 14[/tex]
The rate of change of the area is of 14 cm²/sec.
Item b:
The perimeter is given by:
[tex]P = 2l + 2w[/tex]
Applying implicit differentiation, the rate of change is of:
[tex]\frac{dP}{dt} = 2\frac{dl}{dt} + 2\frac{dw}{dt}[/tex]
Then
[tex]\frac{dP}{dt} = 2(-2) + 2(2) = 0[/tex]
The rate of change of the perimeter is of 0 cm/sec.
Item c:
The diagonal is the hypotenuse of a right triangle in which the sides are the length and the width, then, applying the Pythagorean Theorem:
[tex]d^2 = l^2 + w^2[/tex]
The value of the diagonal is:
[tex]d^2 = 5^2 + 12^2[/tex]
[tex]d^2 = 169[/tex]
[tex]d = \sqrt{169}[/tex]
[tex]d = 13[/tex]
The rate of change is:
[tex]2d\frac{dd}{dt} = 2l\frac{dl}{dt} + 2w\frac{dw}{dt}[/tex]
Then
[tex]26\frac{dd}{dt} = -48 + 20[/tex]
[tex]26\frac{dd}{dt} = -28[/tex]
[tex]\frac{dd}{dt} = -\frac{14}{13}[/tex]
The rate of change of the lengths of the diagonals of the rectangle is of [tex]\mathbf{-\frac{14}{13}}[/tex] cm/sec.
A similar problem is given at https://brainly.com/question/24158553
