Let’s look at this example. We can obviously see a parametric equation because it’s got y’s and x’s
and we’ve got t’s in the equation. The question is asking us to find the first and the second differential.
For the second differential, we’re going to apply the equation that we've just looked at.
We’ll talk a little bit about that equation when we get to actually differentiating it the second time.
So, let’s just do the first part firstly. We’ve got two equations y = 3t +√t
and then our second equation, x = 5t - √t. The first thing, we're going to differentiate this.
Now before I differentiate, remember, we don’t like roots. So, we’re going to write this as 3t + t
to the power of a half. Similarly here, we’re going to write it as 5t - t to the power of a half.
I haven’t done any differentiation yet. I just made it look a little bit easier for myself.
I now need to differentiate them separately and then bring them together. So, dy/dt is going to be 3.
For this term here, you bring the power down and then you decrease the power by 1.
So, when you take away 1 from half, that gives you minus a half. We can tidy this up a little bit.
We can write this as 3 + 1 over 2t to the half which we also know is the same as 3 + 1 over 2√t.
Then just a little extra bit, we can make 2√t as a common denominator, so that the fraction
doesn’t look so complicated. We’ll take 2 √t as a common denominator. So we have to multiply
this 3 by 2√t in order to increase it by that value. So, that gives you 6 √t + 1.
Hopefully, this will make this entire function a little bit easier to use in our calculations.
Dy/dt is done. There's not much more we can do there. Lets have a look at dx/dt.
It's a fairly similar function. Differentiate 5t to give you 5. Differentiate t to the half.
Bring the power down. Then you decrease the power by 1. So, a half minus 1 will give you minus a half.
Again, we can do exactly the same thing as we did in the y function. You can write this as 1 over 2√t.
So, I’ve just taken that down and I’ve changed that to a root. You can take this as a common
denominator 2 √t. And in order to add it to this fraction, you have to times it together.
So, that gives you 10 √t - 1. That is dx/dt. We now find dy/dx so we’ll do dy/dx.
Remember what that was. It was dy/dt multiplied by dt/dx or you can divide it by dx/dt.
To get this, I obviously need to flip this fraction. So rather than dx/dt, I want dt/dx.
So that gives me 2√t. I’m just swapping places. So 2 √t at the top and 10 √t - 1 at the bottom.
By rewriting it in this form, it makes it a little easier for us to multiply it. So, we take dy/dt first
which is 6 √t + 1 over 2 √t. And then I’m going to multiply it with dt/dx. So that gives me 2 √t
over 10 √t - 1. The good thing that happens here is you can cancel this 2 √t with this 2 √t.
I don’t think there’s anything else that simplifies. So we’ve got 6 √t + 1 over 10 √t - 1.
That now is our first differential dy/dx. Once we’ve done this and once we’ve made it as easy
or as simplified as we can because we need to have it simplified for our next differential,
we’re now going to apply our second differential to this. Let’s look at the second differential
of this equation. We’ll start with just getting the definition out. The definition,
as we said earlier, states that the second differential d²y/dx² is the same as dy/dx
but you differentiate this again with respect to dt. And then you divide it with dx/dt.
This looks pretty complicated but it really isn’t when we start to work with functions and numbers.
If it makes it easier for you, you can rewrite this as d/dt of dy/dx. Instead of dividing it by this,
you can times it with dt/dx, similar to what we did with the first differential. It really is up to you
what you prefer to do. So once again, the first part of this equation is saying differentiate dy/dx
with respect to dt, which is what we need to do because our equations have t's in them.
Then the second bit says multiply it with dt/dx. So, if we recall from the previous question
what dy/dx is, we’ve worked this out earlier. We said that that was 6 √t + 1 divided by 10 √t - 1.
We said that dx/dt, so I’m just getting all the things that I need from the previous part,
we've worked this out as well was 10 √t - 1 over 2 √t. We also flipped this in the previous part.
So, we also wrote this as dt/dx. It’s just a matter of flipping it as 2 √t over 10 √t - 1.
So, these things we’ve just done. This is all from the previous part. This dy/dt in here in the definition
is this dy/dx. This dy/dx in here is the same as this dy/dx. We need to differentiate this again.
So, we’re differentiating this expression with respect to t. It makes sense because the expression here
has t’s in it. So, we have to differentiate it. Now, the complication here is that this function
is a quotient rule question because it’s a function being divided by a function.
So, we now have to use the quotient rule firstly to do this part. Let’s just call this the first part
and let’s call this the second part. If we look at the first part, we have to differentiate this.
So this is still the first part. We have to differentiate this with respect to t but unfortunately,
we have to use the quotient rule. Let’s just try that out now. So, the quotient rule has a u
which is 6 √t + 1. It also has a v which is the bottom value which is 10 √t - 1.
So, this is your u and this is your v. We'll differentiate them separately. We can say that u dashed,
now, we need to remember that the t is a power. Before I do that, if I just write this to make it easier,
6t to the half + 1 and then if we write v is 10t to the half - 1. It just makes it a little bit easier
before we start to differentiate. So, the differential of this, bring the power down. The 6 can stay
and decrease the power by 1. So t to the half goes to t to the minus half. This cancels out
to give you 3 over √t. So, that’s your u dashed. We do the same now for v. So, you can say
that v dashed is bring the half down. The 10 can stay. Decrease the power by 1 to give you minus a half
and the constants, obviously this and this disappears. So if we tidy this up, that gives me 5 over √t.
We now apply the quotient rule. The quotient rule, remember, said that in order to differentiate,
udvdudx minus udvdx all over v². So, if we get everything that we need, we have vdudx.
So that times this, we will have 10 √t - 1. That’s being multiplied with 3 over √t.
Then we subtract udvdx, so we’re looking at this function with this function. So, we can rewrite this as
6 √t + 1 and then we multiply this with 5 over √t. We’re getting there. Remember that this, all of this,
obviously don’t forget this part, needs to divide with v². So, we’ve got 10 √t - 1, all squared.
Let’s simplify things a little bit. So, if we carry on here, we can multiply 3 through with this.
We can take √t as a common factor. So, we’ve got 30 √t - 3. Then that’s all over √t.
Then here again, we’ve got 30 √t, leave that in brackets for now, + 5 over √t.
So, we’ve just times it through and then we've still got (10 √t - 1)². It’s all algebra here.
So, just take a deep breath and continue. Hopefully, things get a little bit easier.
At the top, we have √t as a common denominator. We’ve got 30 √t - 3, times it through
with a minus. I've got -30 √t - 5. Then it’s still all being divided by 10 √t - 1, all squared.
There’s one nice little thing here, 30 √t and -30 √t cancels out. We’ve got -3, -5
which gives you -8. This √t is now multiplying with this little expression at the bottom.
We’ve got √t multiplied by 10 √t - 1, all squared. So, I think it’s safe to stop here
and not simplify this anymore. Just remember that what we’re doing, we’re just doing this
point number 1 yet. This is just the first part of it where we are differentiating dy/dx
with respect to d/dt. This usually isn’t that complicated only because we had to use the quotient rule
and we’ve got lots of roots and lots of t’s. We are going through it algebraically.
It’s taking a little bit longer. But it’s still good practice for algebra and we’re almost there.
Back to our definition, we said that d²y/dx² is the same as d/dt of dy/dx, which is what
we’ve been doing just now. You can then either divide it by dx/dt or you could rewrite this as d/dt
of dy/dx and you can multiply it with dt/dx. So, we’re almost at the end. Let’s just get our derivative
or this part or the answer to this part from the previous slide. We said that d/dt of dy/dx
by using the quotient rule gave us -8 over √t(10 √t - 1)². So, that’s this part of the equation.
The only thing left now is to multiply it with dt/dx. So previously, we worked at what dx/dt was.
Our answer for that was 10 √t - 1 over 2 √t. If I want to times it with dt/dx, I just have to flip this.
So, that gives me 2 √t over 10 √t - 1. All I have to do in order to find d²y/dx² is take this function
which is this function and then multiply it with that, so just following the definition.
I have -8 over √t (10√t - 1)². I multiply this with dt/dx to give me 2 √t over 10 √t - 1.
Not much cancels out except these √t’s, √t’s which isn’t so bad. We’ve got 8 times 2 at the top,
so that gives me -16. Then you’ve got 1 bracket and another two brackets. So, you end up
with (10 √t - 1)³. That is now your second differential. It may have put you off the second differential
a little bit doing this question. But remember that this is a harder version of the second differential question.
If we didn’t have those √t’s or didn’t have so many fractions, this would be a lot easier to do.
But it’s good to practice our algebra and good to practice our skills of quotient rules and so on.
But remember that the important thing is that you understand the definition of the second differential.
It is just differentiating dy/dx again with respect to dt and then you just have to divide it by dx/dt.