Question

1)The area of a blot of ink is growing such that after t seconds it's area is given by A= 3t^2+7 cm^2. Calculate the rate if increase of area at t= 5sec.

2. A metallic disc is being heated. It's area at any time t is given by A=5t^2+4t+8. Calculate rate of increase in area at t=3s

Answer 1

1)

Given:

• Area=$3 {t}^{2} + 7$

Need to Find:

• Rate of increase of area at t= 5sec.

Answer:

Rate of increase in area =$\dfrac{dA}{dt}$

putting A= $3 {t}^{2} + 7$

we get,

$\dfrac{dA}{dt} = \dfrac{d(3 {t}^{2} + 7) }{dt}$

$=> \dfrac{dA}{dt} = 3 \times 2t \\ \implies \: \dfrac{dA}{dt} = 6t$

now rate of increase in area at t=5s will be =6×5 ${cm}^{2}$ (putting t=5)

$\large{\boxed{\red{=30 {cm}^{2}}}}$

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2)

Given:

• area= $5 {t}^{2} + 4t + 8$

Need to Find:

• Rate of increase in area at t= 3s.

Answer:

$\dfrac{dA}{dt} = \dfrac{d(5 {t}^{2} + 4t + 8)}{dt} \\ \implies \: \dfrac {dA}{dt} = 10t + 4$

now at t=3s,

Rate of increase in area

= 10×3+4 ${m}^{2} / sec$(putting t=3s)

$\large{\boxed{\red{ = 34 {m}^{2} /sec}}}$