Verify that the following pairs of curves meet orthogonally. a. x^2+y^2=4, x^2=3 y^2 b. x=1-y^

step 1 So section 3 .7, and we're looking at problem number 51. So here they give us two sets of curves

Verify that the following pairs of curves meet orthogonally....

Key Concepts

Slope Product Condition

The condition for two lines to be perpendicular is that the product of their slopes equals -1. In the context of intersecting curves, verifying this condition at the points of intersection confirms that the curves meet at right angles.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent variable is not isolated on one side of the equation. This method allows us to find the derivative (and hence the slope of the tangent) even when x and y are intermixed in the equation.

Orthogonal Intersection

When two curves intersect orthogonally, it means that the angle between their tangent lines at the points of intersection is 90 degrees. This condition is mathematically expressed by the fact that the product of the slopes of the tangent lines at the intersection point is -1.

Tangent Lines

The concept of tangent lines is fundamental when analyzing the behavior of curves at a particular point. A tangent line at a point on a curve represents the instantaneous direction of the curve, and its slope can be determined by differentiating the curve's equation.

Transcript

00:01 So section 3 .7, and we're looking at problem number 51.

00:07 So here they give us two sets of curves, and they ask us to prove that where the curves intersect, they do so orthogonally.

00:16 So i've got x squared plus y squared equal four.

00:22 That's a nice circle, and then x squared is equal to 3 y squared.

00:29 And so i've got to prove that where these things intersect, it happens orthogany.

00:35 So first of all, i've got to find out where they intersect.

00:38 So what i can do is this equation here, x equal 3y squared, let's substitute that value into the first equation.

00:46 So when i do that, i get 3y squared plus y squared is equal to 4, which gives me 4 y squared equal 4, y squared.

01:00 Equal 1, y equal plus or minus 1.

01:05 So two different y values.

01:06 Now let's plug that back into that same equation.

01:10 So i'm gonna get x plus plus or minus 1 squared equal 4.

01:17 So this gives me x squared plus 1 equal 4, x squared equal 3, x equal plus or minus 3.

01:28 So i find that i got four points to worry about.

01:31 Out sorry x squared is plus or minus square root of three so i've got to look at the square root of three one the square root of three negative one negative square root of three one negative square to three minus one i need to prove that both of those curves at each of these four points meet at right angles so i need to go next and figure out then what's going to be the derivative of both of those curves so we've got x squared plus y squared equal four differentiate that with respect to x two x plus two y d y d x equals zero this gives me two y d y d x is equal to minus two x therefore d y d x is equal to minus 2x over 2y, therefore this is just minus x over y.

02:38 Now let's take the other curve, x squared, equal 3y squared, differentiate that with respect to x.

02:46 2x is equal to 6y, d, y, d, y, dx.

02:51 Therefore, d, dx is equal to 2x over 6y, which is just x over 3y.

03:01 So i've got the derivatives of both curves.

03:07 So now what i need to do is just to take those two derivatives.

03:10 So i've got dy, dx is equal to minus x over y.

03:18 And then i'll also have dydx is equal to x over 3y.

03:27 And now i'm just interested in looking at each of these four points.

03:31 So what happens at the square root of 3, 1, the square root of 3, minus 1 minus square root of 3 1 and minus the square root of 3 minus 1 okay so let's evaluate the slope at each of these points so in the first case you're going to get negative square root of 3 over 1 that's just negative square root of 3 and the second case you're going to get the square root of 3 over 3 times 1 which is just one over the square root of three.

04:12 So what i see is these two numbers are negative reciprocals of each other.

04:18 So these two curves, when they meet at that point, they meet at right angles because the slopes of the tangents are perpendicular to each other.

04:26 So just continue on.

04:27 We need to prove this at every point.

04:29 So d, y, dx is going to be minus the square root of three over minus one.

04:34 That's simply the square root of three.

04:37 X over three y is the square root of over negative 3, which is just negative 1 over the square root of 3, and those are, again, negative reciprocals of each other.

04:53 And then negative square root of 3 divided by 1 is negative square root of 3.

05:04 And then if i take, oh, sorry about that...

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