Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Show that the area of the triangle in the first quadrant formed by the tangent line to any point on the curve $y=k / x, k>0$ and the coordinate axes is a constant.

Area $=2 k$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 2

Derivatives Rules 1

Derivatives

Missouri State University

Campbell University

University of Nottingham

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.

03:57

Show that the triangle tha…

02:45

05:06

Find the area of the trian…

00:40

Let $L$ be the tangent lin…

Okay, so we're gonna be doing a fairly simple calculus problem and simple in the sense that the math involved is pretty basic. However, it's a bit complex. There's quite a few steps to it. And it's all to prove that this triangle here, which is formed by the tangent line to the curve Y, equals K over X. Um, has a constant area, so we need to prove that that area is a constant. To do that, let's first find the derivative of our curve that will give us our slope of the tangent line so Y equals K over X. We can rewrite it to B y equals K Times X to the negative ones, prevents us from having to use the quotient rule going forward, we get that y prime by taking the derivative is equal to negative K Times X negative two. We can rewrite this again to give us why prime is equal to negative K all over X squared. That's a piece of information that will need as we go. We'll box that in moving forward. Let's go ahead and rename a couple of the points on our graph. We know for certain that we're gonna need to find our intercepts here because we want to find the area that that encompasses the area of the triangle there. We know here that this is essentially X comma zero, and up here we have zero. Why? These are points. However, we don't know what X and y are. But we do know that on each intercept, the opposite one has to be equal to zero. So finding the area of this triangle recall that the area of a triangle is equal to one half base times height and so we're gonna make this down here. This is our base. Essentially, we're going to rename it that we have our height over here on the Y axis. And now we want to find the we We know for certain that the slope of the tangent line is equal to negative K over X squared. This here being the slope of our tangent line. We know that by taking the derivative of our curve, we get the slope of the tangent line. However, what we need to know now our what our base and height are equal to. So in order to do this, we're gonna split this up into two pieces. We're going to first find the slope of this piece of our tangent line, and then we're going to find another slope of this piece of our tangent line. To do that we're going to take. We were going to recall that M R slope is equal to y Tu minus y one all over x two minus x one and further renaming some of our pieces here. We're going to say that a B is really equal to X y and we know that why is equal to K over X? Because that's what our function is. So we can further rename this to be X comma okay, over X, and that's what that point is. Now our other points for simplicity as we're trying to find our values of our base and height. Let's rename our X intercept to be based zero comma zero, because X whatever distance this is of X is essentially equal to our base and the same for our height. We're going to have zero comma height. This just makes it a little bit easier to follow as we do our work. Alright. So back to determining our slope we're gonna take this. We have. We're going to first use the points that exists right here. Zero comma height and K and X comma K over X. But we're actually going to call it a over B for right now, A little complicated. You'll catch on. So the points that we're working with right now will give us a slope equal to height. So why? To minus and B Y one all over zero x two minus a x one. And we know that this our slope, it has to also be still has to be equal to the slope we found by taking the derivative, which is negative k over X squared, because it's a This tangent line is linear, so that slope is gonna remain the same along the entirety of it. We're just splitting up, splitting it up into two parts so we can again identify the values of our base in height here we want to solve for height. In doing that, we find that going forward we have height minus B is equal to negative K over X squared times a negative, a simplifying that we have. Height is equal to be. We add B to both sides plus a k a over X squared. There's our height, but now we want to make this also a little bit easier to understand because we've got a B, K and X. All of a sudden, there's a lot of variables going on here. So let's simplify this into K over X because that's what we identified. RB is actually equal to plus K X over X squared because remember, a is equal to X. Now that we have this, we can further simplify. We know that these exes are going to cancel, which gives us sorry, my bad. These Xs are going to cancel giving us K over X plus K over X Equalling two K over X and that's our height. Now, to find our base again, we're going to use the slope formula, but we're gonna be using that second portion of our tangent line and those points to determine that base. In doing so, we have em is equal to okay over X minus zero all over X minus base here using our point based comma zero and again X comma K X for that midpoint. And we know also that this is still going to be equal to our constant slope on that tangent line of negative K over X squared, doing some cross multiplication. We're going to get X squared times K over X is equal to negative K times X minus base simplifying. We find that X cancels with that. So we have K X is equal to negative K X distribute, Um, distributing that k out negative k X plus K base. It's the adding K X to both sides. Gives us two K x is equal to K base and then dividing both sides by K, we get a base which is equal to two X. Now that we know what our base and height are, we can go back to our area of a triangle function, which again is one half base times height. Doing that, we end up finding we want in our area. Our triangle, which is equal to our base of two x times are height two K over X, divided by two for that one half. Simplifying this, we see that our X is here. Cancel out and our two's cancel out, which leave us with an area of our triangle equal to two K, which is indeed a constant

View More Answers From This Book

Find Another Textbook

Numerade Educator

02:46

Use the alternate form of the derivative given in Exercise $37,$ to compute …

01:01

Given the data set $(0.0,-12.8),(2.1,-16.9),(3.01,15.1),(3.92,-0.93) (5.01, …

01:09

Normally if we say a limit $L$ exists, we mean that $L$ is a finite number I…

06:07

Show that for any triangle, the medians intersect in a point which is $2 / 3…

02:38

Determine the extrema on the given interval.$$f(x)=x^{2}-2 x+3 \text { o…

Sketch the graph of a function which is decreasing to the left of $x=1,$ inc…

05:01

Use the first derivative to determine where the given function is increasing…

00:47

Sketch the graph of the function defined in the given exercise. Use all the …

03:17

Approximate, using the method of the previous exercise, $f(31.99)$, if $f(x)…

02:32