Baseball players A baseball diamond is a square 90 ft on a side. A player runs from first base t

Baseball players A baseball diamond is a square 90 ft on a...

Key Concepts

Chain Rule

The chain rule is a fundamental differentiation technique used when a variable depends on an intermediate variable that, in turn, depends on time. It allows the conversion of derivatives with respect to one variable into derivatives with respect to time, which is essential in related rates problems.

Trigonometric Relationships

Trigonometric relationships are often used in problems involving angles and distances. By expressing angles in terms of trigonometric functions (such as sine, cosine, or tangent) and using these relationships, one can set up equations that relate the rates at which angles and lengths change over time.

Implicit Differentiation

Implicit differentiation is a technique used when variables are related by an equation not solved explicitly for one variable in terms of another. By differentiating both sides of the equation with respect to time and applying the chain rule, one can relate the rates of change of the variables involved.

Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to time when given the rates of change of related quantities. These problems typically require setting up equations that relate the variables (such as distances, angles, or other measurements) and then differentiating these equations with respect to time to determine the unknown rates.

Transcript

00:01 Hello everyone.

00:02 The question that we are going to solve is something like this.

00:07 Consider a baseball diamond that is actually kind of a square but it's a bit tilted so we are calling it a diamond.

00:15 It's of this kind.

00:20 All sides are equal and the length of the sides is 90 feet.

00:31 Okay.

00:33 Now if this is home, this is home.

00:38 This is first page.

00:39 This is second base this is third base if a player is running from the first base to the second base and he's right here at this instant now the distance from second base to the first base we will call it for example x and the distance of it from the third base let's call it s and what we have to find in part a and also the other thing that is given is it's moving at a speed of 16 feet per second that means x is changing at a speed of 16 feet per second so d x by d t is equal to 16 feet per second now we have to find at what rate the distance of the player with the third base is changing.

01:52 So we actually have to find ds by dt.

01:59 Now let's do that.

02:08 Now as we can see that this triangle here is a right angle triangle.

02:17 So we will apply the pythagoras theorem that will be s square because this angle is the 90 angle.

02:25 S square equal to x square plus and this site is 90 feet so 90 square that is equal to x square plus 8 1 -0 now we will take the derivative of this equation that will give us 2s equal to 2x sorry we are going to take derivative with respect to time d s by d t and 2x d x d x by d t now we want to find this d s by d t when the player has covered this distance which is 30 feet so that means at that is an instant x will be 90 minus 35 feet that is 60 feet now let's solve it and we can find s by plugging in x equal to 60 feet here so s will be 30 square root 13 so now plug in all these values x equal to 60 s equal to this and d x by d t equal to 16 we will plug in all these values here and we will get d s by d t equal to you can just do this is just a easy calculation you can do it on calculator this will be d x by s d x by d t and in numerical value this will be 13 feet per second.

05:07 Now there is one thing to be noted.

05:09 Here in the diagram this x as the player moves the x is decreasing so there should be a negative sign so there will be a negative sign and here overall there will be a negative sign so that means as is decreasing with time.

05:35 Now the second thing that we have to find is if this is angle theta 1 and this is angle theta 2 now we have to find the at what rate are the angles theta 1 and theta 2 change as the time changes as this moves player moves towards the second base you can imagine that angle will be changing so we have to find at what rate these angles will be changing so we have to consider the triangle that we have already highlighted.

06:40 Now let's consider the triangle and from that triangle you can see that cause of theta 1 is equal to 90 divided by s.

06:57 Now we will take the derivative on both sides.

07:01 That will be negative div sine theta 1, d theta 1 divided d t, equal to 90 divided s square, d s by d t.

07:27 Now we know the value of s at when the player has covered 30 feet, and we have to find the rate of change of angle at that instant too.

07:39 So we know the value of s square, we know s by d t at that instance now we have to find sine theta 1 now you can see from the triangle that we can express sine theta 1 in terms of x or s so let's do that that will be d theta 1 divided by d t equal to 90 these negative signs will cancel out and this is s square and sine theta 1 will be x over s d s by d t now s will cancel out this will be 90 over s d s by d t now let us plug in the values and the values are yeah there is x the values are x equal to 60 as in part a s is equal to 30 square root 3 and ds by dt that we found out in last part was negative 32 square root of 13.

09:03 Let's do that.

09:05 You can plug in these values and you will get a numerical answer.

09:10 That will be 8 divided by 65 gradient because this is the unit that we are using per second.

09:28 Oh sorry, this is not the answer...

Please give Ace some feedback