A body cools in a surrounding which is at a c | vedclass

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = -k(	heta - 	heta_0)$,where $k$ is a positive constant

Vedclass · Accepted Answer

$\frac{	an \varphi_2}{	an \varphi_1} = \frac{	heta_2 - 	heta_0}{	heta_1 - 	heta_0}$

Vedclass · Answer

(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = -k(	heta - 	heta_0)$,where $k$ is a positive constant.In the $	heta-t$ graph,the slope of the tangent at any point is $\frac{d	heta}{dt}$. Since the curve is decreasing,the slope is negative. The angle $\varphi$ is the angle made by the tangent with the positive $t$-axis,so the slope is $-	an \varphi$ (as the tangent makes an obtuse angle with the positive $t$-axis,or we consider the magnitude of the slope as $	an \varphi$ where $\varphi$ is the angle with the negative $t$-axis direction).At point $P$ where $	heta = 	heta_2$,the magnitude of the slope is $	an \varphi_2 = k(	heta_2 - 	heta_0)$.At point $Q$ where $	heta = 	heta_1$,the magnitude of the slope is $	an \varphi_1 = k(	heta_1 - 	heta_0)$.Dividing the two equations,we get: $\frac{	an \varphi_2}{	an \varphi_1} = \frac{k(	heta_2 - 	heta_0)}{k(	heta_1 - 	heta_0)} = \frac{	heta_2 - 	heta_0}{	heta_1 - 	heta_0}$.

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