WEBVTT
00:00:02.890 --> 00:00:12.880
Find the equation of the normal to the curve π₯ squared times π¦ squared minus four π₯ plus two π¦ minus 20 is equal to zero at the point one, four.
00:00:13.450 --> 00:00:19.980
The question wants us to find the equation of a normal to a curve defined implicitly at the point one, four.
00:00:20.480 --> 00:00:26.060
To find the equation of a normal, we recall if the tangent to the curve at the point π has a slope of π.
00:00:26.240 --> 00:00:30.980
Then the normal to the curve at the point π has a slope negative one divided by π.
00:00:31.460 --> 00:00:35.460
This is because the tangent and the normal lines are perpendicular to each other.
00:00:35.990 --> 00:00:41.810
In fact, this tells us if the tangent line is horizontal, then the normal line will be vertical, and vice versa.
00:00:42.360 --> 00:00:45.930
So we can use the slope of the tangent to find the slope of our normal.
00:00:45.990 --> 00:00:50.330
To find the slope of our tangent at the point one, four, weβre going to use differentiation.
00:00:50.760 --> 00:00:57.830
We differentiate both sides of our equation with respect to π₯ to help us find an expression for the slope dπ¦ by dπ₯.
00:00:58.160 --> 00:01:07.750
This gives us the derivative of π₯ squared π¦ squared minus four π₯ plus two π¦ minus 20 with respect to π₯ is equal to the derivative of zero with respect to π₯.
00:01:08.220 --> 00:01:10.430
We can differentiate each term separately.
00:01:10.510 --> 00:01:14.450
We know the derivative of zero with respect to π₯ is just equal to zero.
00:01:14.930 --> 00:01:18.740
The derivative of negative 20 with respect to π₯ is also equal to zero.
00:01:18.790 --> 00:01:23.820
And the derivative of negative four π₯ with respect to π₯ is equal to negative four.
00:01:24.580 --> 00:01:28.850
To differentiate the remaining two terms, we recall that π¦ is a function of π₯.
00:01:28.880 --> 00:01:31.390
So we can differentiate these by using the chain rule.
00:01:31.790 --> 00:01:43.150
The chain rule tells us since π¦ is a function of π₯, the derivative of π of π¦ with respect to π₯ is equal to the derivative of π with respect to π¦ multiplied by the derivative of π¦ with respect to π₯.
00:01:43.770 --> 00:01:47.170
We can use this to find the derivative of two π¦ with respect to π₯.
00:01:47.170 --> 00:01:52.460
Itβs equal to the derivative of two π¦ with respect to π¦ multiplied by dπ¦ by dπ₯.
00:01:52.850 --> 00:01:57.080
And we know the derivative of two π¦ with respect to π¦ is just equal to two.
00:01:57.490 --> 00:02:00.990
We see the final term we need to differentiate is the product of two functions.
00:02:00.990 --> 00:02:02.770
So weβll need to use the product rule.
00:02:03.380 --> 00:02:13.130
The product rule tells us the derivative of the product of two functions π’ and π£ with respect to π₯ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.
00:02:13.510 --> 00:02:19.730
So we can use the product rule and the chain rule together to differentiate π₯ squared π¦ squared with respect to π₯.
00:02:19.940 --> 00:02:28.860
Applying the product rule gives us π₯ squared times the derivative of π¦ squared with respect to π₯ plus π¦ squared times the derivative of π₯ squared with respect to π₯.
00:02:29.250 --> 00:02:33.350
We can differentiate π¦ squared with respect to π₯ by using the chain rule.
00:02:33.380 --> 00:02:35.980
Itβs the derivative of π¦ squared with respect to π¦.
00:02:36.110 --> 00:02:39.780
Thatβs two π¦ multiplied by dπ¦ by dπ₯.
00:02:39.970 --> 00:02:44.720
And we can evaluate the derivative of π₯ squared with respect to π₯ as two π₯.
00:02:45.110 --> 00:02:54.000
So we have two π₯ squared π¦ dπ¦ by dπ₯ plus two π₯ π¦ squared minus four plus two dπ¦ by dπ₯ plus zero is equal to zero.
00:02:54.530 --> 00:02:58.280
And we want to find an expression for the slope of our tangents.
00:02:58.280 --> 00:03:02.570
So we need to rewrite this expression to have dπ¦ by dπ₯ as the subject.
00:03:03.130 --> 00:03:07.780
First, weβll subtract two π₯π¦ squared and add four to both sides of the equation.
00:03:08.290 --> 00:03:13.080
Then weβll take out the shared factor of dπ¦ by dπ₯ from the remaining two terms.
00:03:13.670 --> 00:03:20.700
This gives us two π₯ squared π¦ plus two multiplied by dπ¦ by dπ₯ is equal to four minus two π₯π¦ squared.
00:03:21.270 --> 00:03:32.200
Finally, we can divide both sides of our equation by two π₯ squared π¦ plus two to get dπ¦ by dπ₯ is equal to four minus two π₯π¦ squared all divided by two π₯ squared π¦ plus two.
00:03:32.650 --> 00:03:36.440
We want to find the equation of the normal to the curve at the point one, four.
00:03:36.470 --> 00:03:39.610
Thatβs when π₯ is equal to one and π¦ is equal to four.
00:03:39.950 --> 00:03:49.630
So we can substitute π₯ is equal to one and π¦ is equal to four into our expression for dπ¦ by dπ₯ to find the slope of the tangent to the curve at the point one, four.
00:03:50.140 --> 00:03:57.060
Itβs equal to four minus two times one times four squared all divided by two times one squared times four plus two.
00:03:57.590 --> 00:04:01.670
We can evaluate this to get an answer of negative 14 divided by five.
00:04:02.140 --> 00:04:05.160
However, this is the slope of the tangent to the curve at this point.
00:04:05.210 --> 00:04:07.690
We want the slope of the normal to the curve at this point.
00:04:07.790 --> 00:04:11.780
We can do this by finding negative one times the reciprocal of this value.
00:04:12.380 --> 00:04:17.320
The reciprocal of negative 14 divided by five is negative five divided by 14.
00:04:17.350 --> 00:04:21.330
We then multiply this by negative one to just get five divided by 14.
00:04:21.790 --> 00:04:27.920
Since weβve found the slope of our normal line to be five divided by 14, we can write it in the standard form for a line.
00:04:27.960 --> 00:04:31.440
π¦ is equal to five divided by 14π₯ plus π.
00:04:32.040 --> 00:04:36.680
And from the question, we know that our normal line must pass through the point one, four.
00:04:37.030 --> 00:04:43.950
Since our line passes through the point one, four, we can substitute π₯ is equal to one and π¦ is equal to four to find the value of π.
00:04:44.590 --> 00:04:47.500
We subtract five divided by 14 from both sides.
00:04:47.500 --> 00:04:50.970
And we see that π is equal to 51 divided by 14.
00:04:51.350 --> 00:04:56.110
Therefore, by using the fact π is equal to 51 over 14 and then rearranging our equation.
00:04:56.160 --> 00:05:10.410
We have the equation for the normal to the curve π₯ squared π¦ squared minus four π₯ plus two π¦ minus 20 is equal to zero at the point one, four is π¦ minus five π₯ over 14 minus 51 over 14 is equal to zero.