Question Video: Finding the Area of a Plate given Its Rate of Change Using Indefinite Integration Mathematics

Video Transcript

The rate of change in the area 𝐴 of a metallic plate with respect to time due to heating is given by the relation d𝐴 by d𝑡 is equal to 0.036𝑡 squared plus 0.038𝑡, where the area 𝐴 is in square meters and the time 𝑡 is in minutes. Given that 𝐴 is equal to 67 square meters when 𝑡 is equal to eight minutes, find, correct to the nearest two decimal places, the area of the plate just before heating.

In this question, we’re given a problem involving the change in area of a metallic plate due to heating. The area of this metallic plate is given by the function 𝐴 of 𝑡, where 𝐴 is measured in square meters and 𝑡 is measured in minutes. And we need to find the area of the plate just before heating. Since 𝑡 is equal to zero minutes will be when the heating starts, this is just 𝐴 evaluated at zero. However, we can’t just substitute 𝑡 is equal to zero into the function 𝐴 of 𝑡. This is because we’re not given the function 𝐴 of 𝑡. Instead, we’re told that the derivative of 𝐴 with respect to 𝑡 is equal to 0.036𝑡 squared plus 0.038𝑡.

Therefore, we need to use the derivative of this function with respect to time to find the original function. We need to find its antiderivative. And we can recall the indefinite integrals give us the most general antiderivative. In particular, 𝐴 of 𝑡 is going to be equal to the indefinite integral of d𝐴 by d𝑡 with respect to 𝑡. And there is one thing worth pointing out. Since we’re finding the most general antiderivative, this will be only up to a constant of integration 𝐶. To calculate this indefinite integral, let’s start by substituting our expression for d𝐴 by d𝑡 into the integral. We get the indefinite integral of 0.036𝑡 squared plus 0.038𝑡 with respect to 𝑡. And this is the indefinite integral of a polynomial. And we can evaluate this term by term by using the power rule for integration.

This tells us for any real constants 𝑘 and 𝑛, where 𝑛 is not negative one, the integral of 𝑘 times 𝑥 to the 𝑛th power with respect to 𝑥 is equal to 𝑘 times 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus the constant of integration 𝐶. We add one to the exponent of our variable and divide by this new exponent. And we’ll apply this term by term. Let’s start by integrating the first term. We add one to the exponent to get a new exponent of three and then divide by this new exponent. We get 0.036𝑡 cubed over three. And we could add a constant of integration here. However, we can just combine all of the constants of integration into one constant of integration at the end of our expression.

So now let’s integrate our second term. We’ll do this by rewriting 𝑡 as 𝑡 to the first power. We add one to our exponent of 𝑡 to get a new exponent of two and divide by this new exponent. We get 0.038𝑡 squared over two. And then remember, we add our constant of integration 𝐶. And now we can simplify this equation by evaluating the coefficients. We get that 𝐴 of 𝑡 is equal to 0.012𝑡 cubed plus 0.019𝑡 squared plus 𝐶. And now we see we can’t just answer our question by substituting 𝑡 is equal to zero into the equation because we have an unknown value of 𝐶. This means we’re going to need to determine the value of 𝐶. And we can do this by using information we’re given in the question.

We’re told that the area of the metallic plate is equal to 67 square meters when 𝑡 is equal to eight minutes. In terms of our function, this is saying that 𝐴 evaluated at eight is equal to 67. Therefore, we can use this to evaluate our equation at 𝑡 is equal to eight. Substituting 𝑡 is equal to eight, we have 𝐴 evaluated at eight is 67. And this will be equal to 0.012 times eight cubed plus 0.019 times eight squared plus 𝐶. And now we can rearrange this to find the value of 𝐶. If we do this and evaluate, we see that 𝐶 is equal to 59.64. And we can substitute this value of 𝐶 back into our equation for 𝐴 of 𝑡. This gives us that 𝐴 of 𝑡 is 0.012𝑡 cubed plus 0.019𝑡 squared plus 59.64.

And now we can answer our question. We need to substitute 𝑡 is equal to zero into this equation. However, when we do this, our first term will have a factor of zero and our second term will have a factor of zero. This will just leave us with the constant value of 𝐶. Therefore, 𝐴 evaluated at zero is 59.64. But remember, this represents a physical quantity. It’s the initial area of the metallic plate before we start eating. And we’re told that this area is measured in square meters, so we can give this a unit. And this is already given to two decimal places.

Therefore, we’ve shown the initial area of the metallic plate before we started heating, to two decimal places, is 59.64 square meters.