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Question: A metallic disc is being heated. Its area \[A\left( {{\text{ in }}{{\text{m}}^2}} \right)\] at any t...

A metallic disc is being heated. Its area A( in m2)A\left( {{\text{ in }}{{\text{m}}^2}} \right) at any time t(in seconds)t\left( {{\text{in seconds}}} \right) is given by A=5t2+4t+8A = 5{t^2} + 4t + 8. Calculate the rate of increase in area at t=3 st = 3{\text{ s}}.

Explanation

Solution

To solve this question, we will use the concept that the derivative of a function gives the rate of change of function value with the change in argument value. We will first differentiate the given function of area with respect to time. Then we will put the value of t=3 st = 3{\text{ s}} to get the rate of increase in area at t=3 st = 3{\text{ s}}.

Complete step by step answer:
From the question we have;
A=5t2+4t+8A = 5{t^2} + 4t + 8
Now we will calculate the derivative of this function with respect to time. So, differentiating both sides with respect to time we get;
⇒dAdt=d(5t2+4t+8)dt\Rightarrow \dfrac{{dA}}{{dt}} = \dfrac{{d\left( {5{t^2} + 4t + 8} \right)}}{{dt}}
Further expanding the terms in the bracket, we get;
⇒dAdt=d(5t2)dt+d(4t)dt+d(8)dt\Rightarrow \dfrac{{dA}}{{dt}} = \dfrac{{d\left( {5{t^2}} \right)}}{{dt}} + \dfrac{{d\left( {4t} \right)}}{{dt}} + \dfrac{{d\left( 8 \right)}}{{dt}}

Differentiating using the rules of differentiation we get;
⇒dAdt=10t+4+0\Rightarrow \dfrac{{dA}}{{dt}} = 10t + 4 + 0
Because the differentiation of constant is zero.
⇒dAdt=10t+4\Rightarrow \dfrac{{dA}}{{dt}} = 10t + 4
Now we will evaluate the value of the above differential equation at t=3 st = 3{\text{ s}}.
⇒(dAdt)t=3 s=10(3)+4\Rightarrow {\left( {\dfrac{{dA}}{{dt}}} \right)_{t = 3{\text{ s}}}} = 10\left( 3 \right) + 4
∴(dAdt)t=3 s=34\therefore {\left( {\dfrac{{dA}}{{dt}}} \right)_{t = 3{\text{ s}}}} = 34

Therefore, the rate of increase of area at t=3 st = 3{\text{ s}} is 34 m2s−134{\text{ }}{{\text{m}}^2}{s^{ - 1}}.

Note: If in any other question we are given the velocity as a function of time and asked to find the acceleration at a particular time then we can simply differentiate the velocity with respect to time and get the value of acceleration at a particular time. This is because the rate of change of velocity is called the acceleration. On the contrary, if we are given acceleration as a function of time and we are asked to find the velocity as a function of time then we can simply integrate the acceleration function with respect to time. This will give us the velocity as a function of time. These are the physical applications of differentiation and integration.