Problem

A glass of water whose temperature is $50^{\circ}…

06:09

Problem 8 Medium Difficulty

By writing Equation (1.1.7) in the form
$$
\frac{1}{T-T_{m}} \frac{d T}{d t}=-k
$$
and using $u^{-1} \frac{d u}{d t}=\frac{d}{d t}(\ln u),$ derive $(1.1 .8)$

Answer

$T_{m}+c e^{k t}$

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Calculus 2 / BC

Differential Equations and Linear Algebra

Chapter 1

First-Order Differential Equations

Section 1

Differential Equations Everywhere

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Video Transcript

in this problem, we will be solving Newton's law of cooling that says the rate of change with temperature. With respect to time, lower Case T is equal to negative K or case constant times T, which is the temperature at time T minus T sebum, which is a surrounding temperature. In order to solve this differential, equation will start off by dividing both sides of the equation by the factor T minus tisa Ben We'll obtain one divide by T minus tisa bm times The rate of change d t of temperature With respect to time, the little T equals negative K for the next step in solving this differential equation, let's recall that by implicit differentiation, the derivative with respect to do time little case T of the natural law Guerry them of Team Ice. T's of them would be equal to one divide by Team Ice T's a BM then multiplied by the rate of change of temperature. With respect to time d t d. Lower case T. With this equation, we can go off to the next step and express our work as one over or D over DT times Natural log of T minus T sub em equals negative K. Next, we can integrate both sides of this equation to change the expression anti derivative of the derivative D over DT off the natural log rhythm of team Ice. T's mm with respect to T equals the anti derivative of negative K DT now rate to solve both sides of this equation. On the left hand side, we're looking for the derivative and its anti derivative. So you simply obtain the natural log rhythm of T minus tisa them on the right hand side. We obtain negative Katie plus a constant C. Our next step is to solve for the temperature t itself. And we can do that by converting this equation into an exponential equation. We obtain team iced tea. Sebum equals e to the power of negative Katie plus C. At this stage, we can use our laws of exponents to express e to the negative K T plus C as e to the power of C times each of the power of negative Katie. But at the same time, we can add tisa PM to both sides of this equation and this gives us the result that t equals Tisa BM plus let's use a constant C times either the Katie, where lower Casey equals e to the power of Upper Casey and see is any real number.

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