Diffrential calculus; distance problem

In summary, two cars approach an intersection at the same time with the red car 300m east and traveling at 60km/h, and the blue car 450m north and traveling at 75km/h. Using the pythagorean theorem and converting the speeds to m/s, the derivative is found to be 20.2 seconds, which is when the two cars will be closest to each other. It does not matter whether the equation is written as 20.8t - 450 or 450 - 20.8t due to the squares canceling out the difference.
  • #1
jnimagine
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Homework Statement


Two cars approach an intersection at the same time. The red car is 300m east of the intersection and traveling at a speed of 60km/h. The blue car is 450m north of the intersection and traveling at a speed of 75km/h. When are the two cars the closest?


Homework Equations


c^2 = a^2 + b^2


The Attempt at a Solution



I converted the speeds into m/s first then used it in a pythagorean theorem equation and found the derivative.
75km/h = 20.8m/s 60km/h = 16.7m/s
c^2 = (20.8t - 450)^2 + (16.7t - 300)^2
and when i do the derivative i get 20.2. But I was just wondering, does it make a difference whether i do 20.8t - 450 or 450 - 20.8t??
 
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  • #2
Here it doesn't matter because of the squares. 20.8t- 450 is the negative of 450- 20.8t but squaring removes the difference.

But you get "20.2" for what? This problem asks when the two cars will be closest. That requires a time answer. If you mean "The two cars will be closest 20.2 seconds after the starting time (when they are at the given positions)", you must include the "seconds".
 

1. What is differential calculus?

Differential calculus is a branch of mathematics that focuses on the study of rates of change. It deals with the calculation of derivatives, which represent the instantaneous rate of change of a function at a specific point.

2. What is the distance problem in differential calculus?

The distance problem in differential calculus involves finding the distance between two points on a curve or a straight line. This is done by using the derivative of the function to calculate the slope of the tangent line at the two points and then using the distance formula to find the distance between them.

3. How is the distance problem solved using differential calculus?

To solve the distance problem using differential calculus, we first find the derivative of the function at the two points given. Then, we use the slope formula to calculate the slope of the tangent line at each point. Next, we use the distance formula to find the distance between the two points. Finally, we plug these values into the distance formula and solve for the distance.

4. What real-life applications involve the distance problem in differential calculus?

The distance problem in differential calculus has many real-life applications, including calculating the velocity of a moving object, finding the minimum distance between two objects, determining the optimal path for a moving object, and calculating the rate of change in various physical and biological systems.

5. How is the distance problem related to other concepts in differential calculus?

The distance problem is closely related to other concepts in differential calculus, such as the chain rule, optimization, and related rates. These concepts all involve finding derivatives and using them to solve real-world problems involving rates of change.

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