# Question Video: Finding the Equation of a Curve given the Slope of Its Tangent, Which Contains a Root Function, Involving Using Integration by Substitution Mathematics

A curve passes through (0, 1) and the tangent at its point (π₯, π¦) has slope 6π₯ β(8π₯Β² + 1). What is the equation of the curve?

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### 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 weβve 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. Weβll need to use the fact that the curve passes through zero, one to find the particular equation of our curve. So, weβll 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 weβre 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 weβre going to use integration by substitution. Weβll make the substitution π’ is equal to eight π₯ squared plus one. Itβs 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₯ isnβt 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, weβll 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 itβs worth recalling that to integrate a term of the form π₯ to the πth power where π isnβt 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 πΆ.

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 πΆ.

Now, really, what we want to do is find the value of πΆ. So weβre 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 πΆ. 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 πΆ.

Letβs solve for πΆ by dividing through by three-eighths. So eight-thirds is equal to two-thirds plus πΆ. And if we subtract two-thirds from both sides, we see πΆ is equal to six-thirds which is simply two. Replacing πΆ with two, and we do have an equation of the curve, but it doesnβt look very nice. So weβre 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.