Question

The temperature of a well stirred liquid kept open to a cold surrounding is plotted against time. The value of $sec^2\theta_1$ is :-

• A. 1+ 9 $tan^2\theta_1$
• B. 1 + $tan^2\theta_1$
• C. 1+ 3$tan^2\theta_2$
• D. 3+ $tan^2\theta_2$

Answer 1

Answer: (B)

1 + $tan^2\theta_1$

Answer 2

Newton's Law of Cooling states that $\frac{dT}{dt} = -k(T - T_a)$, where $T$ is the temperature of the liquid, $t$ is time, $T_a$ is the ambient temperature, and $k$ is a positive constant. From the graph, we can infer that the ambient temperature $T_a = 20^\circ$C. The slope of the tangent to the temperature-time curve at any point is given by $\frac{dT}{dt}$. The angle $\theta_1$ is associated with the tangent at $T = 35^\circ$C, so $\tan\theta_1 = \frac{dT}{dt}\bigg|_{T=35^\circ\text{C}} = -k(35 - T_a) = -k(35 - 20) = -15k$. The angle $\theta_2$ is associated with the tangent at $T = 30^\circ$C, so $\tan\theta_2 = \frac{dT}{dt}\bigg|_{T=30^\circ\text{C}} = -k(30 - T_a) = -k(30 - 20) = -10k$. We are asked to find the value of $\sec^2\theta_1$. Using the fundamental trigonometric identity, $\sec^2\theta_1 = 1 + \tan^2\theta_1$. This identity directly matches option (B). While we can derive a relationship between $\tan\theta_1$ and $\tan\theta_2$ ($\tan\theta_1 = \frac{-15k}{-10k}\tan\theta_2 = \frac{3}{2}\tan\theta_2$), and thus express $\sec^2\theta_1$ as $1 + \left(\frac{3}{2}\tan\theta_2\right)^2 = 1 + \frac{9}{4}\tan^2\theta_2$, this expression is not among the options. Given that option (B) is a universally true trigonometric identity for $\sec^2\theta_1$, and the question asks for "the value of $\sec^2\theta_1$ is :-", the most appropriate answer is the identity itself, as it correctly represents $\sec^2\theta_1$. The other options represent different mathematical expressions that do not directly equate to $\sec^2\theta_1$ based on the provided information and standard identities.