Question

A metallic disc is being heated. Its area A (in meter squared) at any time t (in second) is given by \[A = 5{t^2} + 4t + 8\] Calculate the rate of increase in area at \[t = 3\;s\]
(A) \[43{m^2}{s^{ - 1}}\]
(B) \[44{m^2}{s^{ - 1}}\]
(C) \[24{m^2}{s^{ - 1}}\]
(D) \[34{m^2}{s^{ - 1}}\]

Answer 1

Hint: We can answer this question very easily by using the concept of differentiation. We are asked to find the rate of increase in the area at a given value of time. To find this we can start by differentiating the formula of area that is given with respect to time. If we differentiate some value with time it is said to be the rate of change of that particular value. So, we can just substitute the time after we get the differential thus, leading us to the required answer.

Complete answer:
We can start by writing down the given data from the question.
The area that is being heated is given by the formula, \[A = 5{t^2} + 4t + 8\]
The time at which we are asked to find the value of rate of change of are is given as \[t = 3\;s\]
To find the rate of change of any value, we differentiate it with respect to time and get the value
That is, for area we find it as, \[\dfrac{d}{{dt}}\left( A \right) = \dfrac{d}{{dt}}\left( {5{t^2}} \right) + \dfrac{d}{{dt}}\left( {4t} \right) + \dfrac{d}{{dt}}\left( 8 \right) = 10t + 4\]
Now that we have the value of the rate of change of area at any time, we can substitute the value of time given to us in the above equation and get the value as, \[10t + 4 = 10 \times 3 + 4 = 44{m^2}{s^{ - 1}}\]
In conclusion, the right answer is option (B) \[44{m^2}{s^{ - 1}}\].

Note: Thermal expansion. As the material heats up the molecules move faster and faster, increasing the average distance between each other. Metals expand when you heat them. When a material is heated, the kinetic energy of that material increases and its atoms and molecules move about more. This area is found using the concept of differentiation.