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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

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 (a) the perimeter, and (b) the area of the rectangle.

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Vishal Baghel | Contributor-Level 10

4 months ago

Since the length x is decreasing and the widthy is increasing with respect to time we have

$\frac{d x}{d t}$ = 5cm/min and $\frac{d y}{d t}$ = A cm/min

(a) The perimeter P of a rectangle will be, P = 2 (x + y)

Q Rate of change of perimeter, $\frac{d P}{d t} = \frac{d^{2}}{d t} (x + y)$

$= 2 (\frac{d x}{d t} + \frac{d y}{d t})$

= 2 ( -5 + 4)

= -2 cm/min

(b) The area A of the rectangle is A = x. y.

Rate of change of area is $\frac{dA}{d t} = \frac{d}{d t} (x \cdot y)$

$= x \frac{d y}{d t} + \frac{d x}{d t} \cdot y .$

= 4x - 5y

So, $\frac{dA}{d t}$ |x = 8ay = 6cm = 4 (8) - 5 (6) = 32 - 30 = 2 cm²/min spherical balloon

Since the length x is decreasing and the widthy is increasing with respect to time we have<math><mrow><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac></mrow></math> = 5cm/min and <math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac></mrow></math> = A cm/min (a) The perimeter P of a rectangle will be, P = 2 (x + y)Q  Rate of change of perimeter, <math><mrow><mfrac><mrow><mi>d</mi><mi>P</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><msup><mrow><mi>d</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo stretchy="false"> (</mo><mi>x</mi><mo>+</mo><mi>y</mi><mo stretchy="false">)</mo></mrow></math><math><mrow><mo>=</mo><mn>2</mn><mrow><mo> (</mo><mrow><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac></mrow><mo>)</mo></mrow></mrow></math>= 2 ( -5 + 4)= -2 cm/min (b) The area A of the rectangle is A = x. y.Rate of change of area is <math><mrow><mfrac><mrow><mi>dA</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>d</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo stretchy="false"> (</mo><mi>x</mi><mo>·</mo><mi>y</mi><mo stretchy="false">)</mo></mrow></math><math><mrow><mo>=</mo><mi>x</mi><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>+</mo><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac><mo>·</mo><mi>y</mi><mo>.</mo></mrow></math>= 4x - 5ySo,  <math><mrow><mfrac><mrow><mi>dA</mi></mrow><mrow><mi>d</mi><mi>t</mi></mrow></mfrac></mrow></math> |x = 8ay = 6cm = 4 (8) - 5 (6) = 32 - 30 = 2 cm2/min spherical balloon

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y (x) = ∫? (2t² - 15t + 10)dt
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Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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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 (a) the perimeter, and (b) the area of the rectangle.

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