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we're now on Part C evaluate the definite integral from zero to ten of H prime of T DT using correct units explain the meaning of the expression in the context of this problem so we know that if we just want to evaluate this definite integral from zero to ten of H prime of T DT this is just the same thing as evaluating the antiderivative of this thing right over here which is just H of T at ten and subtract from that the antiderivative of this evaluated at zero this is the second fundamental theorem of calculus this is exactly how we evaluate definite integrals and when you look at it from this you just see that it's when you evaluate it it just gives us the difference in temperature from 0 minutes to 10 minutes between the temperature of 10 minutes and from that we're taking this we're subtracting from that the temperature at 0 minutes so this is really our change in temperature over over the first over those over those 10 minutes and we can actually evaluate it we know what our temperature was after 10 minutes H of 10 is 43 degrees Celsius so this is 43 right over here and from that we're going to subtract our initial temperature our temperature at 0 minutes which is 66 degrees Celsius we're going to subtract 66 degrees Celsius this gives us negative 23 degrees Celsius so our change in temperature is negative 23 degrees or our temperature has gone down 23 degrees Celsius over the course of the first 10 minutes so that is part that is Part C right over there now let's do Part D Part D D at time T equals 0 biscuits with temperature hundred degrees we removed from an oven so now we're talking about biscuits we started with T now biscuits the temperature of the biscuits at time T is modeled by a differentiable function B for which it is known that B prime of T is equal to this business right over here using the given models at time T equals 10 how much cooler are the biscuits than the T well we know what the temperature of the tea is so we just have to figure out the temperature of the biscuits to figure out how much cooler they are then the T and to figure out the temperature of the biscuits we can essentially just use the exact same idea we can say how much did the we know that the biscuits started off at 100 degrees Celsius and we could say well how much did they cool cool down or what was their change in temperature over the 10 minutes if we know the change in temperature and we know it started at 100 degrees then we can use that information to get what it's temperature is at 10 minutes and then we can answer their question so the change in temperature over those 10 minutes is just the definite integral from 0 to 10 of this business right over here of let me just write it so you see it's the exact same pattern as what we saw over here this was for the cookies now we're talking about the biscuits and this right over here is going to be the definite integral from 0 to 10 B prime of T they give it to us right over here is negative 13 point 8 4 e to the negative 0.173 T DT and now we just evaluate this so we can let's take thee well we can evaluate this what we have to do is because we know that the derivative we know the derivative of e to the ax is a e to the X let me write that the derivative of e to the ax is equal to AE to the ax just from the chain rule derivative the inside is just a and then we multiply that times the derivative of the entire thing and derivative of e to the X is just e to the X or we could say that the integral of e to the ax DX is the same is equal to 1 over a he to the ax plus C and you could take the derivative of this to see that you would get this right over here so using that same idea the antiderivative of this over here just going to be the negative thirteen point eight four and we're going to divide by this by this coefficient on the tee right over here so negative zero point one seven three and you could use your calculator calculators are allowed for this part of the problem but we can do this analytically times e to the negative zero point one seven three times t and we're going to evaluate that we're going to evaluate that from from zero to ten so we're going to evaluate it at ten and subtract from that this thing evaluated at zero so this is going to be let me just let me just factor out this part right over here so it's going to be negative thirteen point eight four over negative zero point one seven three times this evaluate a twenty is 10 so e to the if we multiply this times ten this is negative one point seven three that's when I evaluate it at ten and from that we want to subtract when this is evaluated zero if the exponent here is 0 if T is zero the whole exponent is zero so e to the zero power is just equal to one and now we can get our calculator out to evaluate this so get my t i-85 out and i get i'm just going to evaluate this inner expression right over here so e to the e to the negative one point seven three and from that from that i want to subtract i want to subtract one so that gives me negative 0.8 to two so that's this part in parenthesis and then i want to multiply it times what i have out front so my previous answer times and I won't write the negative just because the negative divided by negative is going to be a positive so it's just thirteen point eight four divided by divided by 0.173 because I didn't write the negatives because they just cancel out and that gives me negative sixty five point eight one seven so this is our change in temperature for the biscuits it's our change in temperature let me write it down negative sixty five point eight one or eight two degrees so this is negative sixty five point eight two degrees Celsius this is our change in temperature this is our change in temp for the biscuits now we know that they started off at a hundred degrees and they went down by 65 point eight two degrees over the ten minutes so let's just we they started at one hundred and then they went down by this previous amount so whoops I want to subtract the answer so minus the answer oh sorry I want to actually add the change in temperature want to be very careful here it went down by sixty five degrees so I really should just say I really should say one hundred one hundred minus sixty five point eight one seven yeah I could keep adding digits if I want to for seven that's enough so after ten the biscuits are 34.1 eight degrees and we already know we already knew what the change in temperature for the or what the temperature of the of the tea is after 10 minutes we already knew as 43 degrees so the biscuits are at 34 degrees the tea is at sorry that biscuits are 34 degrees the tea is at 43 degrees so if we do 43 is where the tea is and we subtract the temperature of the biscuits we see that the the biscuits are eight point eight two degrees cooler eight point eight two degrees Celsius cooler so we could say the biscuits the biscuits are eight point eight two degrees Celsius cooler than the tea and we're done with Part D as well
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