Math Problem Statement

Suppose a particle's position at time t is given by r(t) = (t^2, -2t, ln t). Find the tangential component a_T and the normal component a_N of the acceleration at t = 1.

Solution

We are tasked with finding the tangential and normal components of the acceleration for a particle whose position is given by:

r(t)=(t2,2t,lnt)\vec{r}(t) = \left( t^2, -2t, \ln t \right)

To find the tangential component aTa_T and the normal component aNa_N of the acceleration at t=1t = 1, we will follow these steps:

1. Velocity v(t)\vec{v}(t)

The velocity is the derivative of the position vector:

v(t)=dr(t)dt=(ddt(t2),ddt(2t),ddt(lnt))\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = \left( \frac{d}{dt}(t^2), \frac{d}{dt}(-2t), \frac{d}{dt}(\ln t) \right) v(t)=(2t,2,1t)\vec{v}(t) = (2t, -2, \frac{1}{t})

At t=1t = 1:

v(1)=(2(1),2,11)=(2,2,1)\vec{v}(1) = (2(1), -2, \frac{1}{1}) = (2, -2, 1)

2. Acceleration a(t)\vec{a}(t)

The acceleration is the derivative of the velocity vector:

a(t)=dv(t)dt=(ddt(2t),ddt(2),ddt(1t))\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = \left( \frac{d}{dt}(2t), \frac{d}{dt}(-2), \frac{d}{dt}(\frac{1}{t}) \right) a(t)=(2,0,1t2)\vec{a}(t) = (2, 0, -\frac{1}{t^2})

At t=1t = 1:

a(1)=(2,0,1)\vec{a}(1) = (2, 0, -1)

3. Tangential Component aTa_T

The tangential component of the acceleration is given by:

aT=v(t)a(t)v(t)a_T = \frac{\vec{v}(t) \cdot \vec{a}(t)}{|\vec{v}(t)|}

First, calculate the dot product v(1)a(1)\vec{v}(1) \cdot \vec{a}(1):

v(1)a(1)=(2)(2)+(2)(0)+(1)(1)=41=3\vec{v}(1) \cdot \vec{a}(1) = (2)(2) + (-2)(0) + (1)(-1) = 4 - 1 = 3

Now, calculate the magnitude of v(1)\vec{v}(1):

v(1)=(2)2+(2)2+(1)2=4+4+1=9=3|\vec{v}(1)| = \sqrt{(2)^2 + (-2)^2 + (1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3

Thus, the tangential component is:

aT=33=1a_T = \frac{3}{3} = 1

4. Normal Component aNa_N

The normal component of the acceleration is given by:

aN=a(t)2aT2a_N = \sqrt{|\vec{a}(t)|^2 - a_T^2}

First, calculate the magnitude of a(1)\vec{a}(1):

a(1)=(2)2+(0)2+(1)2=4+1=5|\vec{a}(1)| = \sqrt{(2)^2 + (0)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}

Now, calculate aNa_N:

aN=a(1)2aT2=(5)2(1)2=51=4=2a_N = \sqrt{|\vec{a}(1)|^2 - a_T^2} = \sqrt{(\sqrt{5})^2 - (1)^2} = \sqrt{5 - 1} = \sqrt{4} = 2

5. Conclusion

The tangential component aT=1a_T = 1 and the normal component aN=2a_N = 2.

Thus, the correct answer is (c) aT=1,aN=2a_T = 1, a_N = 2.


Would you like further details or have any questions?

Here are 5 related questions to explore:

  1. How do you find the unit tangent vector in vector calculus?
  2. What is the physical meaning of tangential and normal components of acceleration?
  3. How do you compute curvature from the position vector of a particle?
  4. How does the magnitude of velocity relate to tangential acceleration?
  5. Can the normal component of acceleration ever be zero?

Tip: When solving problems with vectors, always remember to differentiate each component individually and apply the chain rule when necessary.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Tangential and Normal Acceleration
Derivatives

Formulas

Velocity: v(t) = dr(t)/dt
Acceleration: a(t) = dv(t)/dt
Tangential Component: a_T = (v(t) · a(t)) / |v(t)|
Normal Component: a_N = sqrt(|a(t)|^2 - a_T^2)

Theorems

Vector differentiation and the chain rule in calculus

Suitable Grade Level

Grades 11-12 or Undergraduate