Question

a metallic disc is being heated its area A at any time t s is given by A=5t2 +4t +8 . The rate of increase in area at t=3s is

Answer 1

Answer:

$\frac{dA}{dt}=34\:square\:units/sec$

Explanation:

Concept:

If y varies with x, then the rate of change of y with respect to x is $\frac{dy}{dx}$

Given:

$A=5t^2 +4\:t +8$

Differentiate with respect to 't'

$\frac{dA}{dt}=5(2t) +4(1)+0$

$\frac{dA}{dt}=10t+4$

At t=3 sec,

$\frac{dA}{dt}=10(3)+4$

$\frac{dA}{dt}=30+4$

$\frac{dA}{dt}=34\:square\:units/sec$

Answer 2

Answer:

The rate of increase of area at t = 3 s is 34 units

Explanation:

It is given that,

A metallic disc is being heated its area A at any time t is given by :

$A=5t^2+4t+8$ where t is time in seconds

We have to find the rate of increase in area at t = 3 s

Differentiating above equation wrt t as :

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

Putting t = 3 s

The above equation becomes : $10(3)+4=34\ units$

Hence, this is the required solution.