Ex 6.1, 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute & the width y is increasing at the rate of 4 cm/minute. When x = 8 cm & y = 6cm, find the rates of change of (a) the perimeter.Let Length of rectangle = š„
& Width of rectangle = š¦
Given that
Length of rectangle is decreasing at the rate of 5 cm/min
i.e. š š/š š = ā 5 cm/ min
And width of rectangle is increasing at the rate of 4 cm/min
I.e. š š/š š = 4 cm/min
Let P be the perimeter of rectangle
We need to find rate of change of perimeter when š„ = 8 cm & y = 6 cm
i.e. Finding šš/šš” when š„ = 8 cm & š¦ = 6 cm
We know that
Perimeter of rectangle = 2 (Length + Width)
P = 2 (š„ + š¦)
Now,
šš/šš” = (š (2(š„ + š¦)))/šš”
šš/šš”= 2 [š(š„ + š¦)/šš”]
š š·/š š= 2 [š š/š š+ š š/š š]
From (1) & (2) šš„/šš” = ā5 & šš¦/šš” = 4
šš/šš”= 2(ā5 + 4)
šš/šš”= 2 (ā1)
š š·/š š= ā2
Since perimeter is in cm & time is in minute
šš/šš” = ā 2 cm/min
Therefore, perimeter is decreasing at the rate of 2 cm/min
Ex 6.1, 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of (b) the area of the rectangle.Let A be the Area of rectangle
We need to find Rate of change of area when š„ = 8 & š¦ = 6 cm
i.e. š šØ/š š when š„ = 8 cm & š¦ = 6 cm
We know that
Area of rectangle = Length Ć Width
A = š„ Ć š¦
Now,
šš“/šš” = (š(š„. š¦))/šš”
šš“/šš” = šš„/šš” . š¦ + šš¦/šš” . š„.
From (1) & (2) šš„/šš” = ā5 & šš¦/šš” = 4
šš“/šš” = (ā5)š¦ + (4)š„
šš“/šš” = 4š„ ā 5š¦
Putting š„ = 8 cm & š¦ = 6 cm
šš“/šš” = 4 (8) ā 5(6)
Using product rule in x . y
as (u.v)ā = uā v + vā u
šš“/šš” = 32 ā 30
š šØ/š š = 2
Since Area is in cm2 & time is in minute
šš“/šš” = 2 cm2/ min
Hence, Area is increasing at the rate of 2cm2/min

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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