A girl flies a kite at a height of 90 m, the wind carrying the kite horizontally away from her

A girl flies a kite at a height of 90 m, the wind carrying...

A girl flies a kite at a height of $90 \mathrm{~m}$, the wind carrying the kite horizontally away from her at a rate of $7.5 \mathrm{~m} / \mathrm{s}$. How fast must she let out the string when the kite is $150 \mathrm{~m}$ away from her?

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Key Concepts

Pythagorean Theorem

This theorem relates the lengths of the sides of a right triangle, stating that the square of the hypotenuse is equal to the sum of the squares of the other two sides. It is fundamental in problems involving distances and provides a relationship between different dimensions of a system, which is essential in setting up equations for related rates problems.

Related Rates

Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. These problems typically require setting up an equation that connects the different quantities and then differentiating with respect to time.

Implicit Differentiation

Implicit differentiation is a technique used when differentiating equations in which the variables are intermingled rather than isolated on one side. It allows us to find the derivative of one variable with respect to another (commonly time in dynamic problems) even when the relationship between them is not explicitly stated.

Transcript

00:01 Okay, this is a situation we'll be using, we'll be doing related rates with the pythagogram theorem.

00:05 So right here, we could just first plug in the 90s, so we get 90 squared plus x squared equals x squared.

00:15 X being the distance away from the girl, and then the length being the distance away from the kite and to the girl.

00:22 So of course, this would be 80100 plus x squared equals l squared.

00:27 And now we will differentiate both sides with respect to t because of course we're trying to find it through time.

00:32 So we get d t of 8100 plus d over d t of x squared and of course d over d t of the length square.

00:47 So this is zero plus two x d x over d t.

00:53 Of course chain rule and of course we do this the same for this.

00:59 So 2l, dl over dt.

01:05 And we'll be solving this for dl over dt, so we could just divide the 2l.

01:10 So we get dx over dt.

01:14 Of course we have the x and the l, so it would be x equals dl over dt.

01:25 And i will actually just leave, yeah, i know, whoops, we forgot the l on that's it.

01:33 Okay, perfect.

01:35 And of course we could continue over here.

01:37 So let's just rewrite this real quickly.

01:39 So dl, dt equals dx over dt times x over l...

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