CUET UG 2025 — Mathematics Calculus

For an equilateral triangle with side $a$a:

Area $= \dfrac{\sqrt{3}}{4}a^2$=43​​a2

Perimeter $= 3a$=3a

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Given: $\dfrac{dA}{dt} = 4\sqrt{3}$dtdA​=43​ cm²/sec

Differentiating the area formula with respect to time:

$$A = \dfrac{\sqrt{3}}{4}a^2$$A=43​​a2

$$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{4} \times 2a \times \dfrac{da}{dt}$$dtdA​=43​​×2a×dtda​

$$\dfrac{dA}{dt} = \dfrac{\sqrt{3}}{2} \times a \times \dfrac{da}{dt}$$dtdA​=23​​×a×dtda​

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Substituting when $a = 4$a=4 cm:

$$4\sqrt{3} = \dfrac{\sqrt{3}}{2} \times 4 \times \dfrac{da}{dt}$$43​=23​​×4×dtda​

$$4\sqrt{3} = 2\sqrt{3} \times \dfrac{da}{dt}$$43​=23​×dtda​

$$\dfrac{da}{dt} = \dfrac{4\sqrt{3}}{2\sqrt{3}}$$dtda​=23​43​​

$\dfrac{da}{dt} = 2$dtda​=2 cm/sec

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Differentiating the perimeter formula with respect to time:

$$P = 3a$$P=3a

$$\dfrac{dP}{dt} = 3 \times \dfrac{da}{dt}$$dtdP​=3×dtda​

$$\dfrac{dP}{dt} = 3 \times 2$$dtdP​=3×2

$\dfrac{dP}{dt} = 6$dtdP​=6 cm/sec

Therefore, the rate of increase of the perimeter is 6 cm/sec.