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Show that the altitude $h,$ drawn to the hypothenuse of a right triangle (the line drawn from the right angle and perpendicular to the hypothenuse) is the mean proportional between the two segments along the hypothenuse, that is $c_{1} / h=h / c_{2},$ see Figure 15.

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 9

Elements of Geometry

Derivatives

Missouri State University

Campbell University

Harvey Mudd College

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.

02:07

Show that a formula for th…

02:17

Prove that the altitude dr…

00:53

Find the altitude, $h$, of…

01:20

Find the altitude (length …

We're trying to prove here that the height if we go from the Vertex for angle a would equal side a sign be signed c over sign a the height that's right here, going down to side at Well, we could look at the triangle right here, which is a right triangle. And from that we could say that the sign of angle B equals the obstinate signed H oversee. And if we isolate age, we would have side sea times the sign of angle. Okay, but working We go from there. At this point, we are trying to get it to match up here. Remember that we do have the law of signs so we can see we're missing a sign. A In a sign C. There is a law of signs which says that side see over sign angle. See well equal a over sign Anglais. And in that equation, if we were to isolate side, see by multiplying by sign of angle See, we would know that side C is equal to side a times signed c over sign capital. Let now let's go ahead and substitute what we just found in for site C. I would give us HH equals a sign Capital C over sign capital B all times sine b And that is what we started with. We've just rearranged it. We put the sign be on top.

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