Math Problem Statement

Which points in space do the following curves have in common? Do they go through those points at the same time?

vector r(subscript 1) = (t^2, 7t-12, t^2) vector r(subscript 2) = (4t-3, t^2, 5t-6)

Solution

We are given two vector functions r1(t)\mathbf{r_1}(t) and r2(t)\mathbf{r_2}(t) and asked to determine if there are points in space where these curves coincide. In particular, we need to find if there are values of tt such that:

r1(t)=r2(t)\mathbf{r_1}(t) = \mathbf{r_2}(t)

That means we need to solve the following system of equations for tt:

(t2,7t12,t2)=(4t3,t2,5t6)(t^2, 7t - 12, t^2) = (4t - 3, t^2, 5t - 6)

This gives us three separate equations:

  1. t2=4t3t^2 = 4t - 3 (from the first components)
  2. 7t12=t27t - 12 = t^2 (from the second components)
  3. t2=5t6t^2 = 5t - 6 (from the third components)

Step 1: Solve the first equation

t2=4t3t^2 = 4t - 3 Rearranging this gives: t24t+3=0t^2 - 4t + 3 = 0 Factoring: (t3)(t1)=0(t - 3)(t - 1) = 0 So, t=3t = 3 or t=1t = 1.

Step 2: Solve the second equation

7t12=t27t - 12 = t^2 Rearranging: t27t+12=0t^2 - 7t + 12 = 0 Factoring is not easy, so use the quadratic formula: t=(7)±(7)24(1)(12)2(1)=7±49482=7±12t = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2(1)} = \frac{7 \pm \sqrt{49 - 48}}{2} = \frac{7 \pm 1}{2} Thus, t=4t = 4 or t=3t = 3.

Step 3: Solve the third equation

t2=5t6t^2 = 5t - 6 Rearranging: t25t+6=0t^2 - 5t + 6 = 0 Factoring: (t3)(t2)=0(t - 3)(t - 2) = 0 Thus, t=3t = 3 or t=2t = 2.

Step 4: Common solutions

  • From the first equation, we found t=3t = 3 or t=1t = 1.
  • From the second equation, we found t=4t = 4 or t=3t = 3.
  • From the third equation, we found t=3t = 3 or t=2t = 2.

The only common solution among all three equations is t=3t = 3.

Step 5: Check if the curves pass through the same point at t=3t = 3

Let's substitute t=3t = 3 into both vectors:

For r1(3)\mathbf{r_1}(3): r1(3)=(32,7(3)12,32)=(9,2112,9)=(9,9,9)\mathbf{r_1}(3) = (3^2, 7(3) - 12, 3^2) = (9, 21 - 12, 9) = (9, 9, 9)

For r2(3)\mathbf{r_2}(3): r2(3)=(4(3)3,32,5(3)6)=(123,9,156)=(9,9,9)\mathbf{r_2}(3) = (4(3) - 3, 3^2, 5(3) - 6) = (12 - 3, 9, 15 - 6) = (9, 9, 9)

Thus, both curves pass through the point (9,9,9)(9, 9, 9) at t=3t = 3.

Conclusion

The two curves pass through the point (9,9,9)(9, 9, 9) at t=3t = 3, and they go through this point at the same time (i.e., when t=3t = 3).


Would you like further details or clarifications? Here are 5 follow-up questions to deepen your understanding:

  1. How do parametric equations define curves in 3D space?
  2. What are vector-valued functions, and how do they describe motion?
  3. Can curves intersect at different times even if they share common points?
  4. How do you determine whether curves intersect at more than one point?
  5. How does the quadratic formula help solve systems involving vectors?

Tip: When solving systems of vector equations, always solve for each component individually and look for common solutions across all components.

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Math Problem Analysis

Mathematical Concepts

Vector Functions
System of Equations
Quadratic Equations

Formulas

t^2 = 4t - 3
7t - 12 = t^2
t^2 = 5t - 6
Quadratic Formula: t = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Quadratic Formula Theorem

Suitable Grade Level

Grades 10-12