WEBVTT
00:00:01.000 --> 00:00:12.360
The position of a particle is given by the parametric equations π₯ of π‘ equals five square root of two π‘ minus three and π¦ of π‘ equals 12 cube root of π‘ squared plus four.
00:00:12.960 --> 00:00:18.440
Find the slope of the tangent line to the path of the particle at the time π‘ equals two seconds.
00:00:18.880 --> 00:00:24.480
When looking for the slope of a tangent line of a curve, weβre interested in finding the derivative.
00:00:25.200 --> 00:00:29.680
In this question though, weβve been given a pair of parametric equations.
00:00:29.680 --> 00:00:33.840
These are equations for π₯ and π¦ in terms of a third parameter π‘.
00:00:34.280 --> 00:00:43.080
And so we recall that as long as dπ₯ by dπ‘ is not equal to zero, dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘.
00:00:43.640 --> 00:00:46.760
Now, letβs rewrite our equations for π₯ and π¦.
00:00:46.760 --> 00:00:52.000
We can say that π₯ is equal to five times two π‘ minus three to the power of one-half.
00:00:52.000 --> 00:00:56.480
And π¦ is equal to 12 times π‘ squared plus four to the power of one-third.
00:00:56.480 --> 00:01:00.920
π₯ and π¦ are composite functions.
00:01:00.920 --> 00:01:02.320
That is, they are a function of another function.
00:01:02.320 --> 00:01:05.200
And so weβre going to use the chain rule to find their derivatives.
00:01:05.760 --> 00:01:15.760
Now, the chain rule says that if π¦ is some function in π’ and π’ is some function in π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯.
00:01:16.000 --> 00:01:19.080
Now, of course, weβre actually initially looking to find dπ₯ by dπ‘.
00:01:19.080 --> 00:01:21.040
So letβs change our chain rule slightly.
00:01:21.560 --> 00:01:26.400
This time, we can say that dπ₯ by dπ‘ is equal to dπ₯ by dπ’ times dπ’ by dπ‘.
00:01:26.880 --> 00:01:30.120
And weβre going to let π’ be equal to the inner part of our composite function.
00:01:30.120 --> 00:01:35.120
Itβs two π‘ minus three so that π₯ is equal to five π’ to the power of one-half.
00:01:35.720 --> 00:01:39.480
We can differentiate our function π’ with respect to π‘, and we get two.
00:01:39.920 --> 00:01:46.560
Then, to differentiate π₯ with respect to π’, we multiply our entire term by the exponent and reduce that exponent by one.
00:01:46.560 --> 00:01:53.480
So we get a half times five π’ to the power of negative one-half, which is five over two times π’ to the power of negative one-half.
00:01:53.800 --> 00:01:55.760
Now, dπ₯ by dπ‘ is the product of these.
00:01:55.760 --> 00:02:00.160
So itβs two times five over two π’ to the power of negative one-half.
00:02:00.160 --> 00:02:01.520
And we can see that the twos cancel.
00:02:01.920 --> 00:02:09.120
By recognizing that π’ to the power of negative one-half is one over the square root of π’, we write this further as five over the square root of π’.
00:02:09.520 --> 00:02:12.680
But, of course, we were looking to differentiate π₯ with respect to π‘.
00:02:12.680 --> 00:02:17.040
So weβre going to go back to our substitution and replace π’ with two π‘ minus three.
00:02:17.520 --> 00:02:21.800
So dπ₯ by dπ‘ is five over the square root of two π‘ minus three.
00:02:22.160 --> 00:02:24.560
We rewrite our chain rule once again.
00:02:24.560 --> 00:02:26.160
And weβre going to find dπ¦ by dπ‘.
00:02:26.680 --> 00:02:29.920
This time, we let π’ be equal to π‘ squared plus four.
00:02:29.920 --> 00:02:32.360
Once again, that is the inner part of our composite function.
00:02:32.840 --> 00:02:38.040
And π¦ is therefore 12π’ to the power of one-third. dπ’ by dπ‘ is fairly straightforward.
00:02:38.040 --> 00:02:38.920
Itβs two π‘.
00:02:39.280 --> 00:02:47.400
Whereas dπ¦ by dπ’ is a third times 12π’ to the power of negative two-thirds, which is four π’ to the power of negative two-thirds.
00:02:47.760 --> 00:02:50.160
Now, again, dπ¦ by dπ‘ is the product of these.
00:02:50.160 --> 00:02:53.400
Itβs two π‘ times four π’ to the power of negative two-thirds.
00:02:53.680 --> 00:02:57.480
π’ to the power of negative two-thirds is one over π’ to the power of two-thirds.
00:02:57.480 --> 00:02:59.880
So we get eight π‘ over π’ to the power of two-thirds.
00:03:00.160 --> 00:03:03.880
But once again, we want to replace π’ with our original substitution.
00:03:04.200 --> 00:03:09.880
And we therefore find dπ¦ by dπ‘ to be equal to eight π‘ over π‘ squared plus four to the power of two-thirds.
00:03:10.200 --> 00:03:12.400
dπ¦ by dπ₯ is the quotient of these.
00:03:12.400 --> 00:03:15.760
Itβs dπ¦ by dπ‘ divided by dπ₯ by dπ‘.
00:03:15.760 --> 00:03:17.560
And we could look to simplify this a little.
00:03:17.560 --> 00:03:23.840
But remember, weβre trying to find the slope of the tangent line to the path of the particle when π‘ is equal to two.
00:03:24.120 --> 00:03:28.400
And so we substitute π‘ equals two into our expression for the derivative.
00:03:28.400 --> 00:03:35.840
And we get eight times two over two squared plus four to the power of two-thirds divided by five over the square root of two times two minus three.
00:03:36.160 --> 00:03:41.360
This simplifies to 16 over eight to the power of two-thirds divided by five divided by the square root of one.
00:03:41.360 --> 00:03:46.120
But, of course, eight to the power of two-thirds is the cube root of eight squared.
00:03:46.120 --> 00:03:47.680
Now, the cube root of eight is two.
00:03:47.680 --> 00:03:50.040
So we get two squared, which is equal to four.
00:03:50.480 --> 00:03:54.000
The square root of one is one, so we get 16 over four divided by five.
00:03:54.240 --> 00:03:56.320
But 16 divided by four is four.
00:03:56.320 --> 00:04:01.800
So the derivative at π‘ equals two is four divided by five, which is, of course, four-fifths.
00:04:02.120 --> 00:04:10.720
And so, given this pair of parametric equations, the slope of the tangent line to the path of the particle at the time π‘ equals two seconds is four-fifths.