A thin uniform rod of length l cm and a small particle lie on a line separated by a distance of

step 1 Welcome to this lesson. In this lesson we have a thin uniform broad of length L and a small part

A thin uniform rod of length l cm and a small particle lie...

A thin uniform rod of length $l \mathrm{cm}$ and a small particle lie on a line separated by a distance of $a$ cm. If $K$ is a positive constant and $F$ is measured in newtons, the gravitational force between them is $$F=\frac{K}{a(a+l)}$$ (a) If $a$ is increasing at the rate $2 \mathrm{cm} / \mathrm{min}$ when $a=15$ and $l=5,$ how fast is $F$ decreasing? (b) If $l$ is decreasing at the rate $2 \mathrm{cm} / \mathrm{min}$ when $a=15$ and $l=5,$ how fast is $F$ increasing?

A thin uniform rod of length $l \mathrm{cm}$ and a small particle lie on a line separated by a distance of $a$ cm. If $K$ is a positive constant and $F$ is measured in newtons, the gravitational force between them is $$F=\frac{K}{a(a+l)}$$
(a) If $a$ is increasing at the rate $2 \mathrm{cm} / \mathrm{min}$ when $a=15$ and $l=5,$ how fast is $F$ decreasing?
(b) If $l$ is decreasing at the rate $2 \mathrm{cm} / \mathrm{min}$ when $a=15$ and $l=5,$ how fast is $F$ increasing?

Key Concepts

Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to time by relating it to other quantities whose rates of change are known. This typically requires differentiating an equation that relates the physical variables with respect to time using appropriate differentiation rules.

Chain Rule

The chain rule is used in differentiation when a variable is a function of another variable that itself depends on time. It allows one to compute the derivative of a composite function by multiplying the derivative of the outer function with respect to the inner function by the derivative of the inner function with respect to time.

Differentiation of Rational Functions

Differentiating rational functions, which are quotients of functions, involves applying rules such as the quotient rule or rewriting the expression to facilitate differentiation. This approach is essential in problems where the function representing a physical quantity, like force, is given as a ratio of expressions that are functions of time-dependent variables.

Transcript

00:01 Welcome to this lesson.

00:03 In this lesson we have a thin uniform broad of length l and a small particle lyle on a line separated by a distance of a.

00:13 We have a positive constant k and a gravitational force f between them that is giving us k on a times a plus l.

00:27 So the first part we'll find how fast f is decreasing when a is increasing at the rate of 2 centimeters per minute and a is equals to 15 at a point where a is equal to 15 and l equals to 5.

00:49 So a equals to 15, l equals to 5 and we have a that is increasing at the rate so the derivative of a with respect to time is giving us 2.

01:07 Okay this is increasing so the two will be positive.

01:12 So i'll have to write f as the k then i'll have a, a plus l to the power negative 1.

01:29 Okay so here i can have f which is equals to k out then this would be a squared plus a l this is to the power negative 1.

01:50 So for us to find how fast f is decreasing we have to find the derivative of f with respect to time which is actually equals to the derivative of f with respect to a times the derivative of a with respect to time.

02:14 Okay so we already have the derivative of a with respect to time which is 2 so let's find the derivative of f with respect to a.

02:22 So d of f with respect to a is equals to the derivative with respect to a of k then a squared plus a l to the power negative 1.

02:39 So this is equals to negative k, k is capital and i would have a squared plus a l this is to the power negative 2 then we have the derivative of a squared plus a l which is 2 a plus l.

03:12 So the derivative of f with respect to a would be giving us negative k times 2 a plus l then 2.

03:30 This is over a squared plus a l all squared.

03:38 So let's put in the values at a point where a is equals to 15 and l equals to 5.

03:53 Okay so there's no value given for k so we'll have it in terms of k.

04:30 So here we have negative k so this is 30 plus 5 that is 35 so we have negative 35 k.

04:48 So here we would have 25 plus 75 which is 300 then squared which is 90 ,000.

05:01 Okay so here we would have negative 7 on 18 ,000.

05:19 Okay and now we can have the derivative of f with respect to time.

05:25 That would be equals to the derivative of f with respect to the a times the derivative of a with respect to 2.

05:39 So this is equals to negative 7 on 18 ,000.

05:50 Okay and this is 2.

05:59 This is 2 minutes 2 centimeters per minute.

06:05 Okay to fold the small places we have negative 0 .008 k.

06:24 This is k centimeters per minute.

06:32 Alright so that is how fast the f is decreasing.

06:44 Now let's look at the b part where we look at how fast f is increasing given that a is 15 again and l is 5...

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