Two cars leave the same intersection and drive in perpendicular directions. Car A travels east at a speed of 60mihr60\frac{mi}{hr}60hrmi, Car B travels north at a speed of 40mihr40\frac{mi}{hr}40hrmi. Car A leaves the intersection at 2pm2pm2pm, while Car B leaves at 2:30pm2:30pm2:30pm. Determine the rate at which the distance between the two cars is changing at 3pm3pm3pm.

69.62 m i h r 69.62\frac{mi}{hr} 69.62 h r mi

Two cars leave the same intersection and drive in perpendicular directions. Car A travels east at a speed of 60 m i h r 60\frac{mi}{hr} 60 h r mi , Car B travels north at a speed of 40 m i h r 40\frac{mi}{hr} 40 h r mi . Car A leaves the intersection at 2 p m 2pm 2 p m , while Car B leaves at 2 : 30 p m 2:30pm 2 : 30 p m . Determine the rate at which the distance between the two cars is changing at 3 p m 3pm 3 p m .

This problem is pretty tricky, but let's take a look and see how we can go about solving it. We're told 2 cars leave the same intersection and drive in perpendicular directions. Car a travels east at a speed of 60 miles per hour, while car b travels north at a speed of 40 miles per hour. Car a leaves the intersection at 2 PM. Car b leaves the intersection at 2:30 PM. Determine the rate at which the distance between the two cars is changing at 3 PM. Okay. So let's first start by drawing a diagram of the situation. So we have this intersection right here, and we know that car a is going to be traveling east, that's told in this problem. So car a is going to be going in this direction, and it's going in this direction at a at a speed of 60 miles per hour. So we have car a going at 60 miles per hour. Now we're also told that this is traveling perpendicular to car b. Car b is traveling north at a speed of 40 miles per hour. So car b would be traveling in this direction, and it's going to be traveling in this direction at 40 miles per hour. So I'm gonna go ahead and make a little more space for my diagram here because we need to really know what these speeds are. So we see that this is going to be going at 40 miles per hour, and then that's going to be going north. So this is the situation that we're looking at. And what we're asked to do is determine the rate at which the distance between these two cars is changing. Now how can we go about doing that? Well, Well, something that I noticed by looking at this diagram that forms is we have a right triangle situation. So this is a changing geometry where we're dealing with a right triangle, where in essence what's happening is these are going further and further apart, meaning this triangle is going to be growing. So what I'm going to do is write a right triangle, and we're gonna specifically look at this right triangle at the specific time that's indicated. So notice here that we're asked for the 2 cars changing at 3 PM. So we're gonna look at the instance of 3 PM. Now how does 3 PM tell us anything? Well, the way that we can use these times is by looking at the speed and figuring out how far each of these cars has traveled by 3 PM. So if we're looking at 3 PM, first focusing on car a, this is moving at 60 miles per hour. Car a left the intersection at 2 PM. We're now looking at the the intersection at 3 PM and where the 2 cars are. That means this car has been traveling for 1 hour. And if it's moving 60 miles per hour, in 1 hour, it has moved 60 miles. So that's going to be where car a is. Now as for car b, well, car b left the intersection at 2:30 PM. That means that it's been traveling for half an hour. So half of 40 miles per hour, in this case, that's going to be 20 miles. It has moved 20 miles in half an hour. It will move 40 miles in an hour, but at this point, it's only moved 20 miles. So we now have this distance and that distance, and we can find this part of the triangle we'll call z by using the Pythagorean theorem, x squared plus y squared equals z squared. Now in this case, x squared, that's going to be what? Say this is x and this is y. X squared is going to be 60 squared. You won't have plus y squared, which is 20 squared. That's all gonna be equal to z squared, what we're trying to find. 60 squared plus 20 squared, that whole thing just comes out to 4,000. That's gonna be equal to z squared. I can take the square root on both sides of this equation. That's going to give me my z. And the square root of 4,000 is approximately equal to 63.2 miles. So that's going to be our z right here. So we're gonna say this is about 63.2 miles. So now we have these distances, and we have a couple of rates that we're dealing with. We're ultimately trying to find this rate, which I'm going to call dz over dt. And the reason I'm calling it that because that's the rate at which this hypotenuse is changing right here. So that's what we're looking for in this problem. So what I'm going to now do is list out all the variables that we're dealing with in this scenario. So what I can see is that we have an x that we found. The x that we found is 60 miles. We have a y that we found. The y that we found is 20 miles. We also have a z that we just calculated to be 63.2 miles, and then we can take a look at our rates. Well, I can see one of the rates that we have right here is going to be this changing in the x directions, the changing part of the x part of our triangle. So we're gonna see that d x over d t, that's 60 miles per hour. And I can also see right here that we have this port right of the triangle given to us, which is the changing y value. So we have d y over d t. And dy over dt is going to be 40 miles per hour. That's also given to us. And what we're looking for is this rate right here, which is dz over dt. So these are the variables that we have, and this is the variable we're looking for. So what equation is going to relate all of these variables together? Well, I have x's, y's, and z's, and I have rates for x's, y's, and z's. I think it makes sense that the Pythagorean theorem would be perfect for this since that relates all the sides of the triangle together. Now at this point, we've gone ahead and used this diagram. So I'm gonna go ahead and remove it to make some space here, and we're just going to see if we can go through this process for related rates to find our missing rate. So in this case, we have x squared plus y squared equals z squared, and I'm gonna take the time derivative on both sides of this equation. Now we're in the process where we're using the simplest of differentiation and the chain rules to solve this problem. Now using the power rule, the derivative of x squared, that's 2x, then we'll get d x over d t using the chain rule since we're differentiating x with respect to t. Likewise, we can take the derivative of y squared, which is gonna give us 2 y, d y over d t x is with respect to time. That's gonna be equal to z's the derivative of z squared, which would be 2 z, and then we'll get d z over d t. Again, we're taking this time derivative, so we get these time rates in the equation. Now notice we have 2 that shows up everywhere in this equation. I could just divide 2 through on both sides of the equation to get 2 to completely cancel. So it's going to give me the rate x x times the rate for x, and it's going to be plus y times the time rate for y, and it's gonna be equal to z times the time rate for z. So this is what we have so far. Now at this point, I'm trying to find d z over d t. This is the rate that I'm looking for. And since this is the rate that I'm trying to solve for in this problem, well, I'm just gonna divide that z through on both sides of the equation. So that will get the z's to cancel on this side of the equation, giving us that d z over d t, which I'm gonna write over here, is equal to this equation that we have. It's gonna be xdx over dt, that's gonna be plus ydy over dt. That's all going to be divided by z. So that's our equation that we're looking for. So at this point, we've gone ahead and taken this derivative. We know what we're dealing with now. So from here, I just need to plug in the variables that we found in this problem. So doing this, we're trying to find a d z over d t. Now I'm gonna go ahead and plug in the values that we have. So we have x, which we said was 60 miles per hour. And notice everything here is in miles and hours. All the units check out. So here we have the this situation where we have is x here is 60, and that's going to be multiplied by d x over dt, which is also 60. That's gonna be plus y, which we see here is 20, that's given to us in this problem. And or actually, I should say we extracted that information when we drew the triangle, that's multiplied by d y over d t, which right here is 40. And that's going to be divided by z, which you can see is z is 63 point 2. So these are all things you can just plug into your calculator. So you can plug this in as that exactly as you see it, and you should get that your rate is approximately equal to 69.62 on your calculator. So you should get approximately 69.62, and that's going to be in miles per hour. So about 69.6. So if you get an answer around this value, because keep in mind, you are rounding values over here in answers to get them. But if you get around this value, this would be the approximate speed at which these two cars are changing as they move away from each other as this moves in that direction and as this one moves in that direction. So this is how you can solve the problem. That's the solution. Hope you found this video helpful.

Two cars leave the same intersection and drive in perpendicular directions. Car A travels east at a speed of 60 m i h r 60\frac{mi}{hr} 60 h r mi , Car B travels north at a speed of 40 m i h r 40\frac{mi}{hr} 40 h r mi . Car A leaves the intersection at 2 p m 2pm 2 p m , while Car B leaves at 2 : 30 p m 2:30pm 2 : 30 p m . Determine the rate at which the distance between the two cars is changing at 3 p m 3pm 3 p m .

69.62 m i h r 69.62\frac{mi}{hr} 69.62 h r mi

1.58 m i h r 1.58\frac{mi}{hr} 1.58 h r mi

18.99 m i h r 18.99\frac{mi}{hr} 18.99 h r mi

50.41 m i h r 50.41\frac{mi}{hr} 50.41 h r mi

4. Applications of Derivatives

Related Rates

Calculus

4. Applications of Derivatives

Related Rates

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Multiple Choice

Two cars leave the same intersection and drive in perpendicular directions. Car A travels east at a speed of $60\frac{mi}{hr}$ , Car B travels north at a speed of $40\frac{mi}{hr}$ . Car A leaves the intersection at $2pm$ , while Car B leaves at $2:30pm$ . Determine the rate at which the distance between the two cars is changing at $3pm$ .

$69.62\frac{mi}{hr}$

$1.58\frac{mi}{hr}$

$18.99\frac{mi}{hr}$

$50.41\frac{mi}{hr}$

Verified step by step guidance

Define the variables: Let x(t) be the distance traveled by Car A at time t, and y(t) be the distance traveled by Car B at time t. The distance between the two cars at time t is given by the Pythagorean theorem: d(t) = sqrt(x(t)^2 + y(t)^2).

Determine the expressions for x(t) and y(t): Since Car A travels east at 60 mi/hr and starts at 2 pm, x(t) = 60(t - 2) for t >= 2. Car B travels north at 40 mi/hr and starts at 2:30 pm, so y(t) = 40(t - 2.5) for t >= 2.5.

Calculate the time difference: At 3 pm, Car A has been traveling for 1 hour, so x(3) = 60 * 1 = 60 miles. Car B has been traveling for 0.5 hours, so y(3) = 40 * 0.5 = 20 miles.

Differentiate the distance function: To find the rate at which the distance between the cars is changing, differentiate d(t) with respect to t using the chain rule: dd/dt = (1/2)(x(t)^2 + y(t)^2)^(-1/2) * (2x(t)dx/dt + 2y(t)dy/dt).

Substitute the known values: At t = 3, substitute x(3) = 60, y(3) = 20, dx/dt = 60, and dy/dt = 40 into the differentiated equation to find dd/dt, the rate at which the distance between the cars is changing at 3 pm.

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