In this problem. Temperature Capital T is given by the function f of X and T, where X is the distance away from a heater and time is the Ortiz the time after the heater is turned on, Um, we are told that at a time later of five minutes, so At T equals five and at a distance of x meters. I'm sorry at a distance of 3 m or X equals three, um, that the temperature changes at specific rates with respect to time and that with respect to um distance, we're told that the temperature increases at 1.2 degrees Celsius permanent, so that means the rate of change with respect to time, all other things held constant is given by 1.2 meters. I'm sorry, 1.2 degrees Celsius per minute. We're also told that the temperature decreases at two degrees per meter away from the um away from the heater. So that means that the derivative with respect to distance or X is given at negative two. Because, um, with increasing X values, the temperature decreases. So, um, we have to find a We have to find Delta T from this point t equals five and X equals three to appoint later where t equals six minutes and X equals 2.5 m away from the, um away from the heater. In order to do that, we need to approximate or we need to make the assumption that these derivative values here are going to be constant over that range. And that's an approximation we could make because the interval from 5 to 6 and the interval from 2.5 to 3 are small intervals, relatively speaking. So what we do is we say our change and, uh, our change in temperature is going to be our change of temperature. With respect to distance, times are change of distance plus the change of temperature with respect to time, times our change in time. Okay. And once again, this is something that we can. This is a calculation that we can on Lee make under the assumption that the derivatives do not the derivatives do not change over this interval. So we simply plug in the values that we have. Our Delta X is going to be three minus Alright, Sorry. 2.5 minus three. So that's negative in this case. And that makes sense because we're getting closer to the heater from 3 to 2.5. Our value is decreasing so that our Delta X has to be measured in negative. Our, um, derivative with respect to X was also negative. That was negative two. And then we plug in 1.2 degrees Celsius per minute and delta T is from six from five minutes to six minutes. Six minus five is one minute. Alright, so then we perform algebra and we get that. The change in temperature is 2.2 degrees Celsius. Very good.