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Question Video: Applications on Using Indefinite Integration for Finding a Function given Its Derivative Mathematics

A group of laborers are digging a hole, where the rate of change of the volume 𝑉 of the sand removed in cubic meters with respect to the time 𝑑 in hours is given by the relation d𝑉/d𝑑= 𝑑 + 15. Calculate the volume of the sand dug out in 5 hours, rounded to the nearest hundredth.

05:15

Video Transcript

A group of laborers are digging a hole, where the rate of change of the volume 𝑉 of the sand removed in cubic meters with respect to the time 𝑑 in hours is given by the relation d𝑉 by d𝑑 is equal to 𝑑 plus 15. Calculate the volume of the sand dug out in five hours, rounded to the nearest hundredth.

In this question, we’re told that a group of laborers are a digging a hole. And we’re given a relation involving the volume of sand removed with respect to time. The derivative of this volume with respect to time is equal to 𝑑 plus 15, where the volume 𝑉 of sand removed is given in the units cubic meters and the time 𝑑 is given in hours. We need to calculate the volume of sand removed in five hours. We need to round our answer to the nearest hundredth.

To answer this question, we start by noting the volume of sand removed is going to be given by the function 𝑉 of 𝑑. That’s the amount of sand removed after 𝑑 hours. And therefore, since the question is asking us to find the amount of sand removed after five hours, the question wants us to evaluate 𝑉 at five.

However, we’re not given the function 𝑉 of 𝑑. Instead, we’re given d𝑉 by d𝑑. That’s the rate of change in volume with respect to time. And since we know an expression for the derivative of 𝑉 with respect to 𝑑, we need to find an antiderivative of this expression, something which differentiates to give us 𝑑 plus 15. And to do this, we can recall, using indefinite integration, we can find the most general antiderivative. In this case, 𝑉 of 𝑑 will be the indefinite integral of its derivative d𝑉 by d𝑑 with respect to 𝑑. And it’s worth reiterating this is only going to be true up to a constant of integration because we’re finding the most general antiderivative.

To evaluate this, let’s substitute the expression we’re given for d𝑉 by d𝑑 into the equation. We’re told d𝑉 by d𝑑 is 𝑑 plus 15. Therefore, 𝑉 of 𝑑 will be the indefinite integral of 𝑑 plus 15 with respect to 𝑑. We now need to evaluate this integral. We’ll do this term by term by using the power rule for integration, which tells us, for any real value of 𝑛 not equal to negative one, the integral of π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝐢. We add one to the exponent of our variable and divide by the new exponent.

Remember, we’re allowed to evaluate integrals term by term. So we can apply this to evaluate the integral of 𝑑. Well, it’s worth noting we can call the variable in our expression anything. The integral of 𝑑 to the 𝑛th power with respect to 𝑑 will be 𝑑 to the 𝑛 plus one divided by 𝑛 plus one plus 𝐢. Therefore, by rewriting 𝑑 as 𝑑 to the first power, we add one to this exponent to get two and divide by the new exponent. The integral of 𝑑 to the first power is 𝑑 squared over two. And we can add a constant of integration now. However, we’ll get a constant of integration when we integrate each term. And it’s easier to combine these into one constant of integration at the end of our expression.

Now, we need to evaluate the integral of the constant 15. And we can do this by using the power rule for integration. 15 is equal to 15 multiplied by 𝑑 to the zeroth power. However, it’s far easier to recall the derivative of a linear function is the coefficient of 𝑑. In other words, the derivative of 15𝑑 with respect to 𝑑 is 15, which means that 15𝑑 is an antiderivative of 15. Then, finally, we add our constant of integration 𝐢. 𝑉 of 𝑑 is equal to 𝑑 squared over two plus 15𝑑 plus 𝐢.

We want to find 𝑉 evaluated at five. However, there’s a problem. We still have this unknown constant of integration. And to find this value of 𝐢, we’re going to need to know 𝑉 of 𝑑 for some value of 𝑑. And we can do this directly from the question. Remember, 𝑉 of 𝑑 is the amount of sand removed after 𝑑 hours. Therefore, if no time has passed, no sand will have been removed. 𝑉 of zero is equal to zero. And since 𝑑 is equal to zero in this case, this is sometimes called the initial condition.

Now, we can substitute 𝑑 is equal to zero into our function 𝑉 of 𝑑. 𝑉 of zero is zero squared over two plus 15 times zero plus 𝐢. And remember, this needs to be equal to zero since no sand has been removed at time 𝑑 is equal to zero. Zero squared over two is equal to zero, and 15 multiplied by zero is also equal to zero. Therefore, we just have 𝐢 is equal to zero.

We can then substitute 𝐢 is equal to zero into our function 𝑉 of 𝑑. This gives us that 𝑉 of 𝑑 is equal to 𝑑 squared over two plus 15𝑑. And now we can substitute 𝑑 is equal to five into this expression to find the volume of sand removed after five hours. Substituting 𝑑 is equal to five into 𝑉 of 𝑑 gives us 𝑉 evaluated at five is five squared over two plus 15 multiplied by five, which we can evaluate is equal to 87.5.

But we’re not done yet. Remember, the question wants us to round our answer to the nearest hundredth. Therefore, we can add one extra decimal place of accuracy. And finally, remember, this value represents a physical amount. So we should give this units. We’re told the volume 𝑉 is measured in cubic meters provided 𝑑 is measured in hours. So we add the units of cubic meters to our answer, which gives us our final answer.

The volume of sand removed after five hours by the laborers to the nearest hundredth is 87.50 cubic meters.

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