A car leaves an intersection traveling east. Its position t sec later is given by

x = t2 + t
ft. At the same time, another car leaves the same intersection heading north, traveling
y = t2 + 3t ft
in t sec. Find the rate at which the distance between the two cars will be changing 3 sec later. (Round your answer to one decimal place.)

x would actually be 12 not 10

x = t^2 + 1t which is 10 at t=3

y = t^2 + 3t which is 18 at t=3

dx/dt = 2 t + 1 which is 7 at t=3
dy/dt = 2 t + 3 which is 9 at t=3

r^2 = x^2 + y^2 = 434 at t = 3
so at t = 3, r = 20.6

2 r dr/dt = 2 x dx/dt + 2 y dy/dt
20.6 dr/dt = 10(7) + 18(9)
dr/dt = 11.3 ft/s

apple

Well, these cars seem to be going on a date, huh? Let's see how their distance is changing.

We need to find the rate at which the distance between the two cars is changing, so we need to find the derivative of the distance equation.

The distance between the two cars can be found using the Pythagorean theorem:

d² = x² + y²

d² = (t² + t)² + (t² + 3t)²

d² = t⁴ + 2t³ + t² + t⁴ + 6t³ + 9t²

d² = 2t⁴ + 8t³ + 10t²

Now, let's take the derivative of d² with respect to t:

d²/dt = 8t³ + 24t² + 20t

We want to find the rate at which the distance is changing 3 seconds later, so let's plug that in:

t = 3

d²/dt = 8(3)³ + 24(3)² + 20(3)

d²/dt = 8(27) + 24(9) + 60

d²/dt = 216 + 216 + 60

d²/dt = 492

So, the rate at which the distance between the two cars will be changing 3 seconds later is approximately 492 ft/sec. Keep an eye on those cars, they might be racing towards each other!

To find the rate at which the distance between the two cars will be changing 3 seconds later, we need to first find the distance between the two cars at that time and then calculate the derivative of the distance with respect to time.

Let's start by finding the distance between the two cars at t = 3 seconds.

The position of the first car at t seconds is given by x = t^2 + t ft.
Substituting t = 3, we find the position of the first car at t = 3 seconds:
x = (3)^2 + 3 = 12 ft.

The position of the second car at t seconds is given by y = t^2 + 3t ft.
Substituting t = 3, we find the position of the second car at t = 3 seconds:
y = (3)^2 + 3(3) = 18 ft.

Now we can find the distance between the two cars at t = 3 seconds:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
distance = sqrt((18 - 12)^2 + (18 - 12)^2)
distance = sqrt(6^2 + 6^2) = sqrt(36 + 36) = sqrt(72) ≈ 8.49 ft.

Next, we differentiate the distance with respect to time to find the rate at which the distance is changing:

distance = sqrt((x(t) - x(t))^2 + (y(t) - y(t))^2)
Differentiating both sides with respect to t using the chain rule:

d(distance)/dt = d(sqrt((x(t) - x(t))^2 + (y(t) - y(t))^2))/dt
Using the chain rule, we have:

d(distance)/dt = (1/2) * (x(t) - x(t))^2 + (y(t) - y(t))^2)^(-1/2) * (2 * (x(t) - x(t)) * (dx(t)/dt) + 2 * (y(t) - y(t)) * (dy(t)/dt))
Simplifying:

d(distance)/dt = (x(t) - x(t)) * (dx(t)/dt) + (y(t) - y(t)) * (dy(t)/dt)) / distance
Substituting t = 3 seconds:

d(distance)/dt = (12 - 12) * (dx(t)/dt) + (18 - 12) * (dy(t)/dt)) / 8.49
Since the first car is traveling east, dx/dt is the velocity of the first car.
Taking the derivative of x = t^2 + t with respect to t, we find:

dx(t)/dt = 2t + 1
Substituting t = 3 seconds:

dx(t)/dt = 2(3) + 1 = 6 + 1 = 7 ft/s

Since the second car is traveling north, dy/dt is the velocity of the second car.
Taking the derivative of y = t^2 + 3t with respect to t, we find:

dy(t)/dt = 2t + 3
Substituting t = 3 seconds:

dy(t)/dt = 2(3) + 3 = 6 + 3 = 9 ft/s

Now let's substitute all the values into the equation:

d(distance)/dt = (12 - 12) * (7 ft/s) + (18 - 12) * (9 ft/s) / 8.49
d(distance)/dt = 0 * 7 ft/s + 6 * 9 ft/s / 8.49
d(distance)/dt = 54 ft/s / 8.49
d(distance)/dt ≈ 6.36 ft/s

Therefore, the rate at which the distance between the two cars will be changing 3 seconds later is approximately 6.36 ft/s.