Asked by tom
A rectangle is twice as long as it is broad. Find the rate of change of the perimeter when
the breadth of the rectangle is 1m and its area is changing at the rate of 18cm2
/s.
the breadth of the rectangle is 1m and its area is changing at the rate of 18cm2
/s.
Answers
Answered by
bobpursley
length=2*width
area=length*width= width*2width=2 width^2
width= sqrt (area/2)
Perimeter= 2 width + 2 length= 2width+4width=6*width
perimeter=6 sqrt(area/2)
dP/dt=6 d(sqrt(area/2))/dt
take the derivative, and you have it.
area=length*width= width*2width=2 width^2
width= sqrt (area/2)
Perimeter= 2 width + 2 length= 2width+4width=6*width
perimeter=6 sqrt(area/2)
dP/dt=6 d(sqrt(area/2))/dt
take the derivative, and you have it.
Answered by
tom
thanks
Answered by
gony
A rectangle is twice as long as it is broad. Find the rate of change of the perimeter when the breadth of the rectangle is 1m and its area is changing at the rate of 18cm2 /s.
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