Math Problem Statement

Let v(t) = 2t^2 + 3t, a = 0 and b = 4. Find the distance travelled by a particle moving with velocity v(t) during the time interval using Riemann sums and the limit.

Solution

The problem involves determining the distance traveled by a particle with velocity v(t)=2t2+3tv(t) = 2t^2 + 3t over the interval [a,b][a, b], where a=0a = 0 and b=4b = 4. The solution steps involve evaluating a Riemann sum and eventually taking its limit to compute the integral.

Explanation of Each Part

Part (a):

The value of tˉk\bar{t}_k represents the kk-th point of the subinterval in the partition, calculated as: tˉk=a+2k12n(ba)\bar{t}_k = a + \frac{2k - 1}{2n} (b-a) For a=0a = 0 and b=4b = 4, this becomes: tˉk=2k12n4=4(2k1)2n=4k2n\bar{t}_k = \frac{2k - 1}{2n} \cdot 4 = \frac{4(2k - 1)}{2n} = \frac{4k - 2}{n}

Part (b):

The velocity at tˉk\bar{t}_k, v(tˉk)v(\bar{t}_k), is substituted into the velocity function v(t)=2t2+3tv(t) = 2t^2 + 3t: v(tˉk)=v(4k2n)=2(4k2n)2+3(4k2n)v(\bar{t}_k) = v\left(\frac{4k - 2}{n}\right) = 2\left(\frac{4k - 2}{n}\right)^2 + 3\left(\frac{4k - 2}{n}\right)

Part (c):

The summation of v(tˉk)v(\bar{t}_k) across all subintervals is: k=1nv(tˉk)\sum_{k=1}^n v(\bar{t}_k) This is an approximation of the total distance traveled over the interval.

Part (d):

The Riemann sum for the total distance traveled is: k=1nv(tˉk)Δt=bank=1nv(a+2k12n(ba))\sum_{k=1}^n v(\bar{t}_k) \Delta t = \frac{b-a}{n} \sum_{k=1}^n v\left(a + \frac{2k - 1}{2n} (b-a)\right) For a=0,b=4a = 0, b = 4, Δt=40n=4n\Delta t = \frac{4-0}{n} = \frac{4}{n}.

Part (e):

Taking the limit as nn \to \infty gives the exact distance traveled: limnk=1nv(tˉk)Δt=abv(t)dt\lim_{n \to \infty} \sum_{k=1}^n v(\bar{t}_k) \Delta t = \int_a^b v(t) \, dt This integral can be directly calculated: 04(2t2+3t)dt\int_0^4 (2t^2 + 3t) \, dt =[2t33+3t22]04= \left[\frac{2t^3}{3} + \frac{3t^2}{2}\right]_0^4

Step-by-Step Calculation of the Integral

  1. Compute the antiderivative: 2t33+3t22\frac{2t^3}{3} + \frac{3t^2}{2}
  2. Evaluate at t=4t = 4 and t=0t = 0: (2(4)33+3(4)22)(2(0)33+3(0)22)\left(\frac{2(4)^3}{3} + \frac{3(4)^2}{2}\right) - \left(\frac{2(0)^3}{3} + \frac{3(0)^2}{2}\right)
  3. Simplify: 2(64)3+3(16)2=1283+482=1283+24=2003\frac{2(64)}{3} + \frac{3(16)}{2} = \frac{128}{3} + \frac{48}{2} = \frac{128}{3} + 24 = \frac{200}{3}

The total distance traveled is: 2003 units.\frac{200}{3} \text{ units.}

Would you like further breakdowns or additional help? Here are related questions:

  1. What is the significance of using Riemann sums in evaluating integrals?
  2. How can you approximate the integral for a specific nn?
  3. What is the geometrical meaning of v(t)v(t) over time tt?
  4. Can you calculate the distance using midpoint or trapezoidal rules instead?
  5. How do you interpret the integral of velocity in physical terms?

Tip: Always verify your calculations by comparing numerical approximations to symbolic solutions when integrating complex functions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Riemann Sums
Definite Integrals

Formulas

Riemann sum formula
Limit of sums
Definite integral

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12