Parametric Differentiation: Exercise 5

00:01 Our last example is now just practicing the second differential. So, we’re looking at a function where we have to calculate the first and the second differential. Let’s try this out.

00:12 We’ve got the first equation, the first parametric equation given as 5t² + 3t.

00:19 Our x parameter is defined as sin of t. After all this practice, this stage should be fairly easy now.

00:30 We’re going to do dy/dt. So, you’re differentiating this. Bring the power down. That gives you 10t + 3.

00:39 Then we’re also doing dx/dt. We know that sin of t or sin of x differentiates to cos of t.

00:48 Dy/dx, so the definition for our parametric equations is dy/dt multiplied by dt/dx. In order to do that, we need to flip this fraction. So dx/dt, we need to change to dt/dx. Imagine this is cos t over 1.

01:08 If you’ve changed this side, you swapped dt and dx. You also have to do the same on the other side.

01:14 That becomes 1 over cos of t. So, you’re just swapping them as well. When we put this into this equation, dy/dt is 10t + 3. Dx/dt is 1 over cos t. We can rewrite this as 10t + 3 over cos t.

01:38 That’s your dy/dx. The second part then asks you to calculate the second differential of the equation.

01:47 So, if I just define the second differential here, so we’re saying that d²y/dx², if you remember the definition, you have dy/dx. But we had to differentiate this again with respect to dt. Then we divided this with dx/dt. Again, this appears to be a quotient.

02:13 So, you have a function at the top, u and a function at the bottom, v. What we’ll do is we’ll do the first top half of this equation first just to keep things a little bit tidier. Then we’ll divide it as a second step by dx/dt. Let’s differentiate this function again. But in order to do that, we’re going to use the quotient rule because it’s dividing. We’re going to split it into u and v. You can say that u = 10t +3 and v = cos t. Each time now, we’re differentiating with respect to t. So, u dashed is going to be 10.

02:50 V dashed is going to be -sin of t. If we put this all into the quotient rule, so the quotient rule are the gradient here. The top bit of this equation, this part is d/dt of dy/dx.

03:07 Here, we’ll use the quotient rule. So, we’re using vdudx. So, we’re multiplying this with this and this part with that. Let’s just write it here on the side. Dy/dx using the quotient rule is vdudx minus udvdx over v². If we now apply this, we will have 10 multiplied by cos of t minus these two terms, so we’ve got udvdx. We can have 10t + 3 multiplied by -sin t.

03:44 Don't forget to divide this with v². So we have cos t, all squared. Nothing seems to really cancel out but let’s just see if we can simplify this. We have 10 cos of t minus, times these two first before I multiply it with the minus, -10t sin t and then -3 sin t. So, I’m just multiplying it through with a bracket, all over (cos t)². This then gives me 10 cos t + 10t sin t + 3 sin t.

04:28 Sadly, they all look very close to each other but they don’t quite add together because they're all different terms. So, we’ve done this part, this bit here. We’ve done the top part of this equation which is asking you to do d/dt of dy/dx. The only thing left to do then is part two which is dividing it with dx/dt. So, the final differential d²y/dx² involves us taking this entire fraction, 10 cos t + 10t sin t + 3 sin t over cos t, all squared. We’re now dividing it with dx/dt.

05:20 So, either dividing it by dx/dt or you can times it with dt/dx. Remember that it was the same thing.

05:28 You can times it instead with this term here. Rather than divide it with dx/dt, you can times it with dt/dx.

05:36 Now, I'd do this with two steps but I’m running out of space here. So, I’m just going to multiply this with dt/dx. Remember, it’s this term which is dx/dt flipped. So, that’s just going to be 1 over cos of t.

05:50 You can multiply these two terms together because they’re the same. Our final answer for this will be 10 cos t + 10t sin t + 3 sin t all over (cos t)³. It's fairly complicated but we got to use the quotient rule. We got to use trig differentiation. We also did some algebra here.

06:19 Unfortunately, we can’t really use any identities to simplify this a little bit further.

06:23 So this has to be your final answer for d²y/dx².