Math Problem Statement

A curve is given by the following parametric equations. x(t)=t^2sin(t), y(t)=t^2cod(t). Find the arc length, S, of the curve over the interval 0 is less than or equal to t, is less than or equal to 4. Enter the exact value without decimals. S= _______ units

Solution

To find the arc length SS of a parametric curve defined by x(t)x(t) and y(t)y(t) over an interval atba \leq t \leq b, we use the formula:

S=ab(dxdt)2+(dydt)2dtS = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt

Given: x(t)=t2sin(t),y(t)=t2cos(t)x(t) = t^2 \sin(t), \quad y(t) = t^2 \cos(t)

Step 1: Compute dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}

dxdt=ddt(t2sin(t))=2tsin(t)+t2cos(t)\frac{dx}{dt} = \frac{d}{dt}(t^2 \sin(t)) = 2t \sin(t) + t^2 \cos(t)

dydt=ddt(t2cos(t))=2tcos(t)t2sin(t)\frac{dy}{dt} = \frac{d}{dt}(t^2 \cos(t)) = 2t \cos(t) - t^2 \sin(t)

Step 2: Substitute into the Arc Length Formula

S=04(2tsin(t)+t2cos(t))2+(2tcos(t)t2sin(t))2dtS = \int_{0}^{4} \sqrt{\left(2t \sin(t) + t^2 \cos(t)\right)^2 + \left(2t \cos(t) - t^2 \sin(t)\right)^2} \, dt

Step 3: Simplify the Expression

First, expand each squared term:

(2tsin(t)+t2cos(t))2=4t2sin2(t)+4t3sin(t)cos(t)+t4cos2(t)\left(2t \sin(t) + t^2 \cos(t)\right)^2 = 4t^2 \sin^2(t) + 4t^3 \sin(t) \cos(t) + t^4 \cos^2(t)

(2tcos(t)t2sin(t))2=4t2cos2(t)4t3sin(t)cos(t)+t4sin2(t)\left(2t \cos(t) - t^2 \sin(t)\right)^2 = 4t^2 \cos^2(t) - 4t^3 \sin(t) \cos(t) + t^4 \sin^2(t)

Add these expressions:

4t2sin2(t)+t4cos2(t)+4t2cos2(t)+t4sin2(t)=4t2(sin2(t)+cos2(t))+t4(sin2(t)+cos2(t))4t^2 \sin^2(t) + t^4 \cos^2(t) + 4t^2 \cos^2(t) + t^4 \sin^2(t) = 4t^2(\sin^2(t) + \cos^2(t)) + t^4(\sin^2(t) + \cos^2(t))

Since sin2(t)+cos2(t)=1\sin^2(t) + \cos^2(t) = 1, we have:

4t2+t44t^2 + t^4

Thus, the arc length formula becomes:

S=044t2+t4dtS = \int_{0}^{4} \sqrt{4t^2 + t^4} \, dt

Factor out t2t^2:

S=04tt2+4dtS = \int_{0}^{4} t \sqrt{t^2 + 4} \, dt

Step 4: Evaluate the Integral

Let u=t2+4u = t^2 + 4, then du=2tdtdu = 2t dt, or tdt=12dut dt = \frac{1}{2} du.

When t=0t = 0, u=4u = 4. When t=4t = 4, u=20u = 20.

S=12420uduS = \frac{1}{2} \int_{4}^{20} \sqrt{u} \, du

Integrate:

S=12420u1/2du=12[23u3/2]420S = \frac{1}{2} \int_{4}^{20} u^{1/2} du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{4}^{20}

S=13[(20)3/2(4)3/2]S = \frac{1}{3} \left[ (20)^{3/2} - (4)^{3/2} \right]

S=13(20208)S = \frac{1}{3} \left( 20\sqrt{20} - 8 \right)

Therefore, the exact arc length is:

S=202083 unitsS = \frac{20\sqrt{20} - 8}{3} \text{ units}

Would you like me to explain any part of this process in more detail or provide further simplification?

Related Questions:

  1. What is the physical interpretation of arc length in parametric curves?
  2. How does the arc length formula differ for polar coordinates?
  3. What would the arc length be over a different interval, say 0t20 \leq t \leq 2?
  4. Can the integral for arc length always be expressed exactly, or are numerical methods sometimes necessary?
  5. What are some examples of functions where calculating arc length is particularly difficult?

Tip:

When working with parametric curves, always verify if the resulting integrals can be simplified before attempting to solve them directly; this often saves time and effort!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Parametric Equations
Arc Length
Calculus

Formulas

Arc Length Formula: S = ∫ from a to b √[(dx/dt)^2 + (dy/dt)^2] dt
dx/dt = 2t sin(t) + t^2 cos(t)
dy/dt = 2t cos(t) - t^2 sin(t)

Theorems

Pythagorean Identity: sin²(t) + cos²(t) = 1

Suitable Grade Level

College-Level Calculus