Video Transcript
Determine 𝑓 of 𝑡 if 𝑓 triple prime of 𝑡 is equal to negative four root 𝑡 plus five cos 𝑡.
In this problem, we need to determine a function 𝑓 of 𝑡 from its third derivative 𝑓 triple prime of 𝑡. We’ll do this using the process of antidifferentiation, which is the reverse of differentiation. We recall that the general antiderivative of a function 𝑓 of 𝑡 is the function capital 𝐹 of 𝑡 plus 𝐶 such that the first derivative of 𝐹 of 𝑡 is the function lowercase 𝑓 of 𝑡. An antiderivative is not unique, and there are many functions which differ up to a constant which give the same antiderivative. We’ll need to perform the process of antidifferentiation three times. The antiderivative of 𝑓 triple prime of 𝑡 is 𝑓 double prime of 𝑡. That’s the second derivative of 𝑓 of 𝑡. The antiderivative of 𝑓 double prime of 𝑡 is 𝑓 prime of 𝑡, the first derivative of 𝑓 of 𝑡. And the antiderivative of 𝑓 prime of 𝑡 is 𝑓 of 𝑡, the function itself. Each time, we’ll be working with the most general antiderivatives as no specific boundary conditions are given.
Now, the function 𝑓 of 𝑡 is the sum of two terms, and we recall that antiderivatives are linear. The antiderivative of a sum is the sum of the antiderivatives of each term. We can therefore find the antiderivative of each term separately and combine the constants of antidifferentiation to find the general antiderivative of the entire function. The first term in the function 𝑓 triple prime of 𝑡 can be rewritten using laws of exponents as negative four 𝑡 to the power of one-half. And so we see that the function 𝑓 triple prime of 𝑡 consists of a power term and a trigonometric term. We need to recall general rules for finding antiderivatives of each of these.
First, using the power rule of differentiation, we know that the derivative of 𝑡 to the power of 𝑎 plus one over 𝑎 plus one is 𝑡 to the power of 𝑎, provided 𝑎 is a real constant not equal to negative one. So the most general antiderivative of 𝑡 to the power of 𝑎, where 𝑎 must again not be equal to negative one, is 𝑡 to the power of 𝑎 plus one over 𝑎 plus one plus 𝐶.
We also know that the antiderivative of a constant 𝑎 multiplied by a function lowercase 𝑓 of 𝑡 is that same constant 𝑎 multiplied by the antiderivative capital 𝐹 of 𝑡. In other words, if a derivative is multiplied by a constant, the antiderivative is multiplied by the same constant, and vice versa. For the trigonometric term, we recall that the derivative of sin 𝑡 is cos 𝑡. The derivative of cos 𝑡 is negative sin 𝑡. It follows then that the derivative of negative sin 𝑡 is negative the cos of 𝑡. And the derivative of the negative cos 𝑡 is sin 𝑡.
To find antiderivatives, we need to go the other way around this circle of differentiation, remembering to add constants of antidifferentiation. So let’s begin finding 𝑓 double prime of 𝑡 then. The antiderivative of negative four 𝑡 to the power of one-half using the reverse of the power rule is negative four 𝑡 to the power of three over two over three over two. And we’ll include one constant of antidifferentiation at each stage when we’ve dealt with both terms. Dividing by three over two is equivalent to multiplying by two over three. So the constant here simplifies to negative eight over three.
For the second term, the antiderivative of cos 𝑡 is sin 𝑡. So the antiderivative of five cos 𝑡 is five sin 𝑡. We’ll also include one constant of antidifferentiation, which we’ll call 𝐶 hat, and it will become clear later why we’ve done this. So we found that 𝑓 double prime of 𝑡 is negative eight 𝑡 to the power of three over two over three plus five sin 𝑡 plus 𝐶 hat.
Let’s continue to find 𝑓 prime of 𝑡. The antiderivative of negative eight 𝑡 to the power of three over two over three is negative eight 𝑡 to the power of five over two over three times five over two. Dividing by five over two is equivalent to multiplying by two-fifths. So the first term is equivalent to negative 16𝑡 to the power of five over two over 15. The antiderivative of sin 𝑡 is negative cos 𝑡. So the antiderivative of the second term is negative five cos 𝑡. We’ll need to include a constant of antidifferentiation 𝐷. But we must also remember to find the antiderivative of our previous constant of antidifferentiation 𝐶 hat.
If we think of 𝐶 hat as 𝐶 hat multiplied by 𝑡 to the power of zero, which we can do because 𝑡 to the power of zero is one, then we find that the antiderivative of 𝐶 hat is 𝐶 hat 𝑡. So we have that 𝑓 prime of 𝑡 is equal to negative 16𝑡 to the power of five over two over 15 minus five cos 𝑡 plus 𝐶 hat 𝑡 plus 𝐷. We can then proceed in the same way to find 𝑓 of 𝑡. The antiderivative of negative 16𝑡 to the power of five over two over 15 is negative 16𝑡 to the power of seven over two over 15 multiplied by seven over two, which simplifies to negative 32𝑡 to the power of seven over two over 105. And remember, we’ll include one constant of antidifferentiation for the entire function.
The antiderivative of negative five cos 𝑡 is negative five sin 𝑡. The antiderivative of 𝐶 hat 𝑡 is 𝐶 hat 𝑡 squared over two. As before, we must remember to find the antiderivative of the constant 𝐷, which is 𝐷𝑡. And then we include one constant of antidifferentiation 𝐸 for the entire function. We found then an expression for 𝑓 of 𝑡. But we can simplify it slightly. In the third term, 𝐶 hat over two is just another constant, which we’ll call 𝐶. So by using the process of antidifferentiation three times, we found that the most general function 𝑓 of 𝑡, which has the given third derivative, is 𝑓 of 𝑡 equals negative 32𝑡 to the power of seven over two over 105 minus five sin 𝑡 plus 𝐶𝑡 squared plus 𝐷𝑡 plus 𝐸 for constants 𝐶, 𝐷, and 𝐸.
