Video Transcript
A group of workers are digging a hole. Initially, the depth of the hole was 14 inches. The depth of the hole increases at the rate modeled by the function 𝑅 of 𝑡 equals seven minus 𝑡 times the natural logarithm of 𝑡 plus one over 𝑡 plus four, where 𝑡 is the time in hours and 𝑅 of 𝑡 is measured in inches per hour. How deep will the whole be after six hours of digging, approximated to the nearest inch?
This question is about the depth of a hole. We are given that, initially, the depth of the hole was 14 inches. And we’re also given the rate at which this depth increases. The depth of the hole increases at the rate modeled by the function 𝑅 of 𝑡 equals seven minus 𝑡 times the natural logarithm of 𝑡 plus one over 𝑡 plus four. What we are required to find is how deep the hole will be after six hours of digging. So we’re given the initial depth and the rate of change of depth. And we want to find the final depth. The key to solving this question is to think of depth as a function of time 𝑡.
We let 𝑓 of 𝑡 be the depth, in inches, at time 𝑡 measured in hours. We can then write the given and required quantities in terms of this function 𝑓. So the initial depth is the depth at time 𝑡 equals zero. So this is 𝑓 of zero. And we know that its value is 14. This function 𝑅 of 𝑡 gives the rate of increase of the depth and so is the rate of change of the depth function 𝑓 of 𝑡. In other words, it’s 𝑓 prime of 𝑡, the derivative of 𝑓 of 𝑡. We just read off its value. And what we’re required to find is the depth at time 𝑡 equals six hours. In other words, we’re required to find the value of 𝑓 of six.
Now, how do we find 𝑓 of six given 𝑓 of zero and the derivative of 𝑓? We use part of the fundamental theorem of calculus, which says that the definite integral between two numbers 𝑎 and 𝑏 of the derivative of a function is equal to the difference between the values of that function at 𝑏 and 𝑎. This is sometimes called the total change theorem because the value of 𝑓 of 𝑏 minus 𝑓 of 𝑎 is the total change in the value of the function from 𝑎 to 𝑏. And the theorem states that this total change is equal to this integral here, which we can interpret as the continuous analogue of adding up all the little changes in the function 𝑓 from 𝑎 to 𝑏. For our purposes, it’s best to rearrange this equation to get 𝑓 of 𝑏 alone on the left-hand side. We do this by adding 𝑓 of 𝑎 to both sides and swapping the sides.
Now, we’re ready to apply this to our function. Remember, we’re looking for the value of 𝑓 of six. So 𝑏 is six. And the value of the function that we’re given is that of 𝑓 of zero. So 𝑎 is zero. We see then, with these values, that 𝑓 of six equals 𝑓 of zero plus the definite integral from zero to six of 𝑓 prime of 𝑡 d𝑡. This is true for any differentiable function 𝑓 on the domain from zero to six. But we know that, for our function, 𝑓 of zero is 14. And we also have an expression for the derivative of 𝑓. Substituting then, we find that 𝑓 of six is 14 plus the definite integral from zero to six of seven minus 𝑡 times the natural logarithm of 𝑡 plus one over 𝑡 plus four d𝑡.
Now, we just have to evaluate this integral. And you might think that, to do this, we have to find an antiderivative of the integrand. But this is not the case. In the question, we’re not asked for an exact closed-form value for 𝑓 of six, the depth of the hole after six hours of digging, but only asked to this value to the nearest inch. In other words, we only need 𝑓 of six correct to the nearest whole number. This means we don’t need an exact closed-form value for our integral. We only need an approximate numerical value, which our calculator can give us.
How you find this value using a calculator will depend on what model of calculator you have. But on a TI-84, you press math and then nine. This puts you into definite integral mode. And then, you can just enter this integral much as you see it. My calculator gives me the value of 38.52944153. Adding this to 14, we find the approximate value 52.529 and so on for 𝑓 of six. Now, remember that 𝑓 of six is the depth in inches of this hole after six hours. And we’re only asked to this value to the nearest inch. So we round 𝑓 of six to 53. And because we’re not asked for the value of 𝑓 of six in the question, but for the depth of the hole after six hours of digging, we need to add the unit back. Our answer is therefore 53 inches.
This answer seems reasonable given the context. The initial depth was 14 inches. And they dug a further 38.5 or so inches in the six hours. And so we end up with the hole of depth about 53 inches. The trick to answering this question was to think of the depth as a function of time and then apply the fundamental theorem of calculus. This gave us an answer involving an integral. But as this integral was a definite integral and as we only needed an approximate value, we could use our calculator to evaluate the integral.
If you were troubled by the fact that the function 𝑓 is positive despite representing a depth below ground, don’t worry. We could’ve said 𝑓 of zero to be negative 14. But then, in this case, we would also have needed to set 𝑓 prime of 𝑡 to be negative 𝑅 of 𝑡, in order to get the right answer. With this choice, we would’ve found that 𝑓 of six was negative 53. And so our answer for the depth of the hole after six hours would be the same, 53 inches.
