### Video Transcript

A curve passes through zero, one and the tangent at its point π₯, π¦ has slope six π₯ times the square root of eight π₯ squared plus one. What is the equation of the curve?

The key to answering this question is to spot that weβve been given information about the slope of the tangent at a given point. We remember that we can find a general equation for the slope of a tangent to the curve by differentiating the equation of the curve. In other words, in this case, dπ¦ by dπ₯ is equal to six π₯ times the square root of eight π₯ squared plus one. Or we can write this alternatively as six π₯ times eight π₯ squared plus one to the power of one-half.

Now, by recalling the fact that integration and differentiation are the reverse processes of one another, we find that we can find an expression for π¦ by integrating our expression for dπ¦ by dπ₯ with respect to π₯. Now, essentially, this will give us a general equation. Weβll need to use the fact that the curve passes through zero, one to find the particular equation of our curve. So, weβll integrate six π₯ times eight π₯ squared plus one to the power of one-half with respect to π₯.

Now, you might be wondering how on Earth weβre going to integrate this function. Well, we have the product of two functions, one of which is a composite function. The key here is to spot that the derivative of part of our composite function is equal to a scalar multiple of another part of our function. In other words, the derivative of eight π₯ squared plus one is some multiple of six π₯. This is a good indication to us that weβre going to use integration by substitution. Weβll make the substitution π’ is equal to eight π₯ squared plus one. Itβs the inner part of our composite function.

Next, we differentiate π’ with respect to π₯, remembering that to differentiate a power term, we multiply the entire term by the exponent then reduce the exponent by one. So we get two times eight π₯ to the power of one or two times eight π₯ which is 16π₯. Of course, the derivative of one is zero. So, dπ’ by dπ₯ is 16π₯. Now, dπ’ by dπ₯ isnβt a fraction. But in this process, we treat it a little like one. And we say that this is equivalent to a sixteenth dπ’ equals π₯ dπ₯. Now, the whole point of doing this is we can now replace eight π₯ squared plus one with π’. And we can replace π₯ dπ₯ with a sixteenth dπ’. So this might look like the integral of six times π’ to the power of one-half times a sixteenth dπ’.

But of course, we can rewrite this a little bit more nicely as six over 16 times π’ to the power of one-half dπ’. And then, weβll take out that constant factor of six over 16 or three over eight. And so π¦ is equal to three-eighths times the indefinite integral of π’ to the power of half dπ’. And here itβs worth recalling that to integrate a term of the form π₯ to the πth power where π isnβt equal to negative one, we add one to the power and then divide by that new value. So, the integral of π’ to the power of one-half is π’ to the power of three over two divided by three over two. And of course, we need that constant of integration πΆ.

We know that dividing by a fraction is the same as multiplying by the reciprocal of that fraction. And so, π’ to the power of three over two divided by three over two is the same as two-thirds π’ to the power of three over two. Now, remember, we want an equation for the curve, so that will be π¦ in terms of π₯. So we go back to our substitution π’ is equal to eight π₯ squared plus one. And so we have π¦ equals three-eighths times two-thirds of eight π₯ squared plus one to the power of three over two plus πΆ.

Now, really, what we want to do is find the value of πΆ. So weβre going to go back to the very first bit of information about the curve and the fact that it passes through zero, one. In other words, when π₯ is equal to zero, π¦ is equal to one. And so, we substitute these values in. One equals three-eighths times two-thirds eight times zero squared plus one to the power of three over two plus πΆ. Now, eight times zero squared plus one is just one, and one to the power of three over two is still one. So we get one equals three-eighths times two-thirds plus πΆ.

Letβs solve for πΆ by dividing through by three-eighths. So eight-thirds is equal to two-thirds plus πΆ. And if we subtract two-thirds from both sides, we see πΆ is equal to six-thirds which is simply two. Replacing πΆ with two, and we do have an equation of the curve, but it doesnβt look very nice. So weβre going to distribute three-eighths across our parentheses. Three-eighths times two-thirds is one-quarter, and three-eighths times two is three-quarters. And so the equation of our curve is π¦ equals a quarter times eight π₯ squared plus one to the power of three over two plus three-quarters.