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Show that the tangent line to any point on the circle $x^{2}+y^{2}=a^{2}$ is always perpendicular to the radius drawn to the point of tangency.

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

01:37

For the circle $x^{2}+y^{2…

03:41

Find an equation of the ta…

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Tangent Line Use implicit …

01:41

Suppose that the point $\l…

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Show that any two tangent …

01:58

Show that the equations of…

for this problem. We're going to examine a circle and our circle has an A. The equation of X squared plus y squared equals a squared so as our radius going out a all the way around. And what we want to show is if I draw a radius toe any point, we're going to call this point HK. So it's any random point on this circle If I do a tangent line at that point, the radius and the tangent line meet at perpendicular there, perpendicular to each other. They meet it right angles. So we want to prove that that is the case. So let's take a look at our to First of all, I have the radius. I'll call that are just so I could talk about it and I have the tangent line. We'll just call that line. L So what do we know? Let's look at the radius first. What is the slope off that radius that we drew? Well, I know it begins at the origin 00 and ends in HK, so if I want the slope of that radius, that's gonna be rise overrun. The change in why Over the change in X So the change in why it's k over zero. The change in my X coordinates is H over zero. We're just k over h. So that's the slope of the radius. Now what about the slope of line L the tangent line? Well, for that, let's use the derivative. I have the equation X squared plus y squared equals a squared. If I take the derivative of this, remember, this is gonna be implicit differentiation. Why is a function of X So I have two x plus two y and because of the chain rule, I need to put d y DX in there. Yeah, is a constant. So the derivative there is just zero. Now let's solve for a slope d y dx. I have two y d u I d X equals negative two x Um, I can divide both sides by two to get rid of those twos there. Which means that d y d X equals negative x over y. So that is the slope anywhere around this circle. But I want to find it at a particular point. What? It's the slope at the point of Tange Insee at the point HK. All that means I'm gonna be negative. H over K. Well, look at those two slopes. One is K over H and the other is negative. H over K. They are negative, reciprocal of each other. And that is how perpendicular lines are related. Their slopes are negative, reciprocal of each other. We have that case here, so that shows that these two lines are indeed perpendicular to each other.

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