Vector Calculus.

Vector Calculus

Parametric equations – 5.6 Jacplus
Connections to the Study Design: AOS 4 – Vectors Vector Calculus Position vector as a function of time 𝑟 𝑡 , and sketching the corresponding path given 𝑟 𝑡 , including circles, ellipses and hyperbolas in Cartesian and parametric forms

Parametric Equations Parameter: Path traced out by the particle is defined in terms of a third variable, 𝑡. In a two dimensional case the path is described by two parametric equations, as both the 𝑥− and 𝑦− coordinates depend upon the parameter, 𝑡. 𝑥=𝑥 𝑡 𝑦=𝑦(𝑡) Recap: The position vector of the particle is given by 𝑟 𝑡 =𝑥𝑖+𝑦𝑗, where 𝑖 and 𝑗 are unit vectors in the 𝑥 and 𝑦 directions. Also called a vector function of the scalar real variable 𝑡, where 𝑡 is called the parameter, and often represents time. If we can eliminate the parameter from the two parametric equations and obtain an equation of the form 𝑦=𝑓(𝑥), this is called an explicit relationship or the equation of the path. It may not be possible to obtain an explicit relationship, but often an implicit relationship of the form 𝑓(𝑥,𝑦)=0 can be formed. Careful consideration needs to be given to the possible values of 𝑡, which then specify the domain (the 𝑥−values) and the range (the 𝑦−values) of the equation of the path.

Parametric Equations - Background
A locus is a set of points traced out in the plane, satisfying some geometrical relationship. The path described by a moving particle forms a locus and can be described by a Cartesian equation. Note: The Cartesian equation does not tell us where the particle is at any particular time. The path traced out by the particle can be defined in terms of a third variable – in this case we will use the variable 𝑡 as the parameter. 𝑥=𝑥 𝑡 𝑦=𝑦(𝑡) A position vector is given by 𝑟(𝑡)=𝑥𝑖+𝑦𝑗, where 𝑖 and 𝑗 are unit vectors in the 𝑥 and 𝑦 directions, this is also called the vector equation of the path. If we can eliminate the parameter from these two parametric equations and obtain an equation of the form 𝑦=𝑓(𝑥), then this is called an explicit relationship and is the equation of the path. At times it may be unable to obtain an explicit relationship but an implicit relationship can be obtained in the form of 𝑓(𝑥,𝑦)=0. Either way, the relationship between 𝑥 and 𝑦 is called the Cartesian equation of the path.

Worked Example 19 Given the vector equation 𝑟 𝑡 = 𝑡−1 𝑖+2 𝑡 2 𝑗, for 𝑡≥0, find and sketch the Cartesian equation of the path, and state the dome and range.

Eliminating the parameter
Worked Example 20 Methods to use: Direct substitution Trigonometric formulas Given the vector equation 𝑟 𝑡 = 2+5 cos 𝑡 𝑖+ 4 sin 𝑡 −3 𝑗 for 𝑡≥0, find and sketch the Cartesian equation of the path, and state the domain and range.

Parametric Representations and Sketching parametric curves
Worked Example 22 The parametric representation of a curve is not necessarily unique. To sketch parametric curves, CAS calculators can be used to draw the Cartesian equation of the path from the two parametric equations, even if the parameter cannot be eliminated. Show that the parametric equations 𝑥 𝑡 = 3 2 𝑡+ 1 𝑡 and 𝑦 𝑡 =2 𝑡− 1 𝑡 where 𝑡∈𝑅\ 0 represent the hyperbola 𝑥 2 9 − 𝑦 =1.

Position vectors as functions of time – 13.2 Jacplus
Position vector as a function of time 𝑟 𝑡 , and sketching the corresponding path given 𝑟 𝑡 , including circles, ellipses and hyperbolas in Cartesian and parametric forms

Worked example 1 - Cambridge
For a vector function 𝑟 𝑡 =𝑥 𝑡 𝑖+𝑦 𝑡 𝑗: The domain of the Cartesian relation is given by the range if the function 𝑥(𝑡). The range of the Cartesian relation is given by the range of the function 𝑦(𝑡). In the Example 1b, the domain of the corresponding Cartesian relation is the range of the function 𝑥 𝑡 =1− cos 𝑡 , which is 0,2 . The range of the Cartesian relation is the range of the function 𝑦 𝑡 = sin 𝑡 , which is −1,1 . Note: the Cartesian equation 𝑦 2 =− 𝑥 2 +2𝑥 can be written as 𝑥−1 2 +𝑦=1; it is the circle with centre 1,0 and radius 1. Find the Cartesian equation for the graph represented by each vector function: 𝑟 𝑡 = 2−𝑡 𝑖+ 3+ 𝑡 2 𝑗, 𝑡 ∈𝑅 𝑟 𝑡 = 1− cos 𝑡 𝑖+ sin 𝑡 𝑗, 𝑡 ∈𝑅

Worked Examples - Cambridge
Find the Cartesian equation of each of the following. State the domain and range and sketch the graph of each of the relations. 𝑟 𝑡 = cos 2 𝑡 𝑖+ sin 2 𝑡 𝑗, t∈𝑅 𝑟 𝑡 =𝑡𝑖+(1−𝑡)𝑗, t∈𝑅 For each of the following, state the Cartesian equation, the domain and range of the corresponding Cartesian relation and sketch the graph: 𝑟 𝜆 = 1−2 cos 𝜆 𝑖+3 sin 𝜆 𝑗 𝑟 𝜆 =2 sec 𝜆 𝑖+ tan 𝜆 𝑗

Closest approach Worked Example 1 - Jacplus
Given the position vector of a particle, 𝑟 𝑡 =𝑥𝑖+𝑦𝑗, where 𝑖 and 𝑗 are unit vectors in the 𝑥 and 𝑦 directions, it is possible to find the positon or coordinates of the particle at a given value of t. It is also possible to find the value of 𝑡 and the coordinates when the particle is closest to the origin. A particle moves so that its vector equation is given by 𝑟 𝑡 = 3𝑡−4 𝑖+ 4𝑡−3 𝑗 for 𝑡≥0. Find the distance of the particle from the origin when 𝑡=2. Show that the distance of the particle from the origin at any time 𝑡 is 𝑟(𝑡) = 25 𝑡 2 −48𝑡+25 Hence find the closest distance of the particle from the origin.

Collision problems Worked Example 2 - JacPlus
Two particles move so that their position vectors are given by 𝑟 𝐴 𝑡 = 3𝑡−8 𝑖+ 𝑡 2 −18𝑡+87 𝑗 and 𝑟 𝐵 𝑡 = 20−𝑡 𝑖+ 2𝑡−4 𝑗 for 𝑡≥0. Find: When and where the particles collide. Where their paths cross. The distance between the particles when 𝑡=10.

Position vectors as a function of time – 12B Cambridge

Consider the following scenarios:
A particle travelling at a constant speed along a circular path with radius length 1 𝑢𝑛𝑖𝑡 and centre 𝑂. The path is represented in Cartesian form as: 𝑥,𝑦 : 𝑥 2 + 𝑦 2 =1 If the particle starts at the point (1,0) and travels anticlockwise, taking 2𝜋 units of time to complete one circle, then its path is represented in parametric form as: 𝑥,𝑦 :𝑥= cos 𝑡 𝑎𝑛𝑑 𝑦= sin 𝑡 , 𝑓𝑜𝑟 𝑡≥0 This is expressed in vector form as: 𝑟 𝑡 = cos 𝑡 𝑖 + sin 𝑡 𝑗 where 𝑟(𝑡) is the position vector of the particle at time 𝑡. The graph of a vector function is the set of points determined by the function 𝑟(𝑡) as 𝑡 varies. In two dimensions, the 𝑥− and 𝑦− axis are used. In three dimensions, three mutually perpendicular axes are used. It is best to consider the 𝑥− and 𝑦−axes as in the horizontal plane and the 𝑧−axis as vertical and through the point of intersection of the 𝑥− and 𝑦− axes.

Information from the vector function
The vector function gives much more information about the motion of the particle than the Cartesian equation of its path. For example, the vector function 𝑟 𝑡 = cos 𝑡 𝑖+ sin 𝑡 𝑗, 𝑡≥0, indicates that: At time 𝑡=0, the particle has position vector 𝑟 0 =𝑖. That is, the particle starts at (1,0). The particle moves with constant speed on the curve with equation 𝑥 2 + 𝑦 2 =1. The particle moves in an anticlockwise direction The particle moves around the circle with a period of 2𝜋, ie. It takes 2𝜋 units of time to complete one circle. The vector function 𝑟 𝑡 = cos 2𝜋𝑡 𝑖+ sin 2𝜋𝑡 𝑗 describes a particle moving anticlockwise around the circle with equation 𝑥 2 + 𝑦 2 =1, but this time the period is 1 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑡𝑖𝑚𝑒. The vector function 𝑟 𝑡 = −cos 2𝜋𝑡 𝑖+ sin 2𝜋𝑡 𝑗 again describes a particle moving around the unit circle, but the particle starts at (−1,0) and moves clockwise. Worked Example 4 - Cambridge Sketch the path of particle where the position at time t is given by 𝑟 𝑡 =2𝑡𝑖+ 𝑡 2 𝑗, where 𝑡≥0

Notes: The equation 𝑟 𝑡 =𝑡𝑖 𝑡 2 𝑗, 𝑡≥0, gives the same Cartesian path, but the rate at which the particle moves along the path is different. Motion in two dimensions: When a particle moves along a curve in a plane, its position is specified by a vector function of the form 𝑟 𝑡 =𝑥 𝑡 𝑖+𝑦 𝑡 𝑗 Motion in three dimensions: When a particle moves along a curve in three-dimensional space, its position is specified by a vector function of the form 𝑟 𝑡 =𝑥 𝑡 𝑖+𝑦 𝑡 𝑗+𝑧 𝑡 𝑘 If 𝑟 𝑡 =−𝑡𝑖 𝑡 2 𝑗, 𝑡≥0, the again the Cartesian equation is 𝑦= 𝑥 2 4 , but 𝑥≤0. Hence the motion is along the curve shown and in the direction indicated.

An object moves along a path where the position vector is given by 𝑟 𝑡 = cos 𝑡 𝑖+ sin 𝑡 𝑗+2𝑘, 𝑡≥0 Describe the motion of the object. The motion of two particles is given by the vector functions 𝑟 1 𝑡 = 2𝑡−3 𝑖+ 𝑡 𝑗 and 𝑟 2 𝑡 = 𝑡+2 𝑖+ 7𝑡 𝑗 where 𝑡≥0. Find: The point at which the particles collide The points at which the two paths cross The distance between the particles when 𝑡=1

Vector Calculus – 13.3 and 13.5 jacplus
Connections to the Study Design: AOS 4 – Vectors Vector Calculus Differentiation and anti-differentiation of a vector function with respect to time and applying vector calculus to motion in a plan including projectile and circular motion

Vector Calculus The derivative of r w.r.t 𝑡 is denoted by 𝑟 and is defined by: 𝑟 𝑡 = lim ℎ→0 𝑟 𝑡+ℎ −𝑟(𝑡) ℎ provided that this limit exits. The vector 𝑟 𝑡 points along the tangent to the curve at 𝑃, in the direction of increasing 𝑡. Note: The derivative of a vector function 𝑟(𝑡) is also denoted by 𝑑𝑟 𝑑𝑡 or 𝑟 ′ (𝑡). Additionally: A unit tangent vector at 𝑡=𝑎 is denoted by 𝑠 = 𝑟 (𝑎) 𝑟 (𝑡) . Consider the curve defined by 𝑟(𝑡) Let 𝑃 and 𝑄 be points on the curve with positions vectors 𝑟(𝑡) and 𝑟(𝑡+ℎ) respectively. Then 𝑃𝑄 =𝑟 𝑡+ℎ −𝑟(𝑡) It follows that 1 ℎ 𝑟 𝑡+ℎ −𝑟(𝑡) is a vector parallel to 𝑃𝑄 . As ℎ→0, the point Q approaches P along the curve.

Rules for differentiating vectors
Derivative of a constant vector Derivative of a sum or difference of vectors If 𝑐 is a constant vector, that is a vector which does not change and is independent of t, then 𝑑𝑐 𝑑𝑡 =0. Note that 𝑑𝑖 𝑑𝑡 = 𝑑𝑗 𝑑𝑡 = 𝑑𝑘 𝑑𝑡 =0 The sum or difference of two vectors can be differentiated as the sum or difference of the individual derivatives. That is, 𝑑 𝑑𝑡 𝑎+𝑏 = 𝑑𝑎 𝑑𝑡 + 𝑑𝑏 𝑑𝑡 and 𝑑 𝑑𝑡 𝑎−𝑏 = 𝑑𝑎 𝑑𝑡 − 𝑑𝑏 𝑑𝑡 Using these rules, if 𝑟 𝑡 =𝑥 𝑡 𝑖+𝑦 𝑡 𝑗, then 𝑑𝑟 𝑑𝑡 = 𝑑𝑥 𝑑𝑡 𝑖+ 𝑑𝑦 𝑑𝑡 𝑗 Simply put, to differentiate a vector we merely differentiate each component using the rules for differentiation.

Derivative of a vector functions
Two Dimensions 𝒓 𝒕 =𝒙 𝒕 𝒊+𝒚 𝒕 𝒋, only if 𝒙(𝒕) and 𝒚(𝒕) are differentiable. Three Dimensions 𝒓 𝒕 =𝒙 𝒕 𝒊+𝒚 𝒕 𝒋+𝒛 𝒕 𝒌 First Derivative: 𝑟 𝑡 = 𝑑𝑥 𝑑𝑡 𝑖+ 𝑑𝑦 𝑑𝑡 𝑗 𝑟 𝑡 = 𝑥 𝑖+ 𝑦 𝑗 Second Derivatives: 𝑟 𝑡 = 𝑑 2 𝑥 𝑑𝑡 2 𝑖+ 𝑑 2 𝑦 𝑑 𝑡 2 𝑗 𝑟 𝑡 = 𝑥 (𝑡)𝑖+ 𝑦 (𝑡)𝑗 First Derivative: 𝑟 𝑡 = 𝑑𝑥 𝑑𝑡 𝑖+ 𝑑𝑦 𝑑𝑡 𝑗+ 𝑑𝑧 𝑑𝑡 𝑘 𝑟 𝑡 = 𝑥 𝑖+ 𝑦 𝑗+ 𝑧 𝑘 Second Derivatives: 𝑟 𝑡 = 𝑑 2 𝑥 𝑑𝑡 2 𝑖+ 𝑑 2 𝑦 𝑑 𝑡 2 𝑗+ 𝑑 2 𝑧 𝑑 𝑡 2 𝑟 𝑡 = 𝑥 (𝑡)𝑖+ 𝑦 (𝑡)𝑗+ 𝑧 (𝑡)𝑘

Properties of the derivative of a vector function
𝑑 𝑑𝑡 𝑐 =0, where 𝑐 is a constant vector 𝑑 𝑑𝑡 𝑘𝑟 𝑡 =𝑘 𝑑 𝑑𝑡 𝑟 𝑡 , where 𝑘 is a real number 𝑑 𝑑𝑡 𝑟 1 𝑡 + 𝑟 2 𝑡 = 𝑑 𝑑𝑡 𝑟 1 𝑡 + 𝑑 𝑑𝑡 𝑟 2 𝑡 𝑑 𝑑𝑡 𝑓 𝑡 𝑟 𝑡 =𝑓 𝑡 𝑑 𝑑𝑡 𝑟 𝑡 + 𝑑 𝑑𝑡 𝑓 𝑡 𝑟 𝑡 , where 𝑓 is a real-valued function

A curve is described by the vector equation 𝑟 𝑡 =2 cos 𝑡 𝑖 +3 sin 𝑡 𝑗 . Find: 𝑟 (𝑡) 𝑟 𝑡 Find the gradient of the curve at the point (𝑥,𝑦), where 𝑥=2 cos 𝑡 and 𝑦= 3sin 𝑡 Worked Example 12 A curve is described by the vector equation 𝑟 𝑡 = sec 𝑡 𝑖+ tan 𝑡 𝑗, with 𝑡∈ − 𝜋 2 , 𝜋 2 . Find the gradient of the curve at the point (x,y), where 𝑥= sec 𝑡 and 𝑦= tan 𝑡 . Find the gradient of the curve where 𝑡= 𝜋 4 . Worked Example 7 Find 𝑟 (𝑡) and 𝑟 𝑡 if 𝑟 𝑡 =20𝑡𝑖+ 15𝑡−5 𝑡 2 𝑗. Worked Example 8 Find 𝑟 (𝑡) and 𝑟 𝑡 if 𝑟 𝑡 = cos 𝑡 𝑖− sin 𝑡 𝑗+5𝑡𝑘. Worked Example 9 If 𝑟 𝑡 =𝑡𝑖+ 𝑡− 𝑗, find 𝑟 (𝛼) and 𝑟 𝛼 , where 𝑟 𝛼 =𝑖+𝑗. Worked Example 10 If 𝑟 𝑡 = 𝑒 𝑡 𝑖+ 𝑒 𝑡 − 𝑗, find 𝑟 (𝛼) and 𝑟 𝛼 , where 𝑟 𝛼 =𝑖+𝑗

Worked Example 3 - Jacplus
Find a unit tangent vector to 𝑟 𝑡 = 𝑒 3𝑡 𝑖+ sin 2𝑡 𝑗 at the point where 𝑡=0.

Derivative summary Velocity vector Speed Acceleration vector
Because 𝑟(𝑡) represents the position vector, 𝑣 𝑡 = 𝑑𝑟 𝑑𝑡 = 𝑟 (𝑡) represents the velocity vector. Note the single dot above r indicates the derivative w.r.t time. Furthermore, if 𝑟 𝑡 =𝑥 𝑡 𝑖+𝑦 𝑡 𝑗, then 𝑟 𝑡 = 𝑥 𝑡 𝑖+ 𝑦 𝑡 𝑗. The speed of a moving particle is the magnitude of the velocity vector. The speed at time 𝑡 is given by 𝑟 (𝑡) = 𝑥 𝑦 2 . If the particle has a mass of 𝑚, then the magnitude of the momentum acting on the particle is given by 𝑝=𝑚 𝑟 (𝑡) . Since 𝑣 𝑡 = 𝑑𝑟 𝑑𝑡 = 𝑟 (𝑡) represents the velocity vector, differentiating again w.r.t 𝑡 gives the acceleration vector. The acceleration vector is given by 𝑎 𝑡 = 𝑑 𝑑𝑡 𝑟 𝑡 = 𝑟 𝑡 = 𝑥 𝑡 𝑖+ 𝑦(𝑡 )𝑗. Note that the two dots above the variables indicate the second derivative w.r.t time.

Worked example 4 - jacplus
A particle spirals outwards so that its position vector is given by 𝑟 𝑡 =𝑡 cos 𝑡 𝑖 + 𝑡 sin 𝑡 𝑗 for 𝑡≥0. Find the velocity vector. Show that the speed of the particle at time 𝑡 is 1+ 𝑡 2 and hence find the speed when 𝑡= 3𝜋 4 . Find the acceleration vector.

Worked example 7 - Jacplus
The position vector, 𝑟(𝑡), of a golf ball at a time 𝑡 seconds is given by 𝑟 𝑡 =15𝑡𝑖+ 20𝑡−4.9 𝑡 2 𝑗 for 𝑡≥0, where the distance is in metres, 𝑖 is a unit vector horizontally forward and 𝑗 is a unit vector vertically upwards above ground level. Find when the golf ball hits the ground Find where the golf ball this the ground Determine the initial speed and angle of projection Find the maximum height reached Show that the golf ball travel in a parabolic path.

Antidifferentiation The constant Vector
When integrating a function, always remember to include the constant of integration, which is a scalar. When integrating a vector function w.r.t a scalar, the constant of integration is a vector. This follows since if 𝑐 is a constant vector, then 𝑑 𝑑𝑡 𝑐 =0.

Integration of vector functions
Integrating an Acceleration Vector to gives a Position Vector Integrating a Velocity Vector w.r.t time gives a Position Vector 𝑟 𝑡 = 𝑥 𝑡 𝑑𝑡 𝑖+ 𝑦 𝑡 𝑑𝑡 𝑗=𝑥 𝑡 𝑖+𝑦 𝑡 𝑗+𝑐 Note: that an initial condition must be given in order for us to be able to find the constant vector of integration. Given the acceleration vector 𝑎 𝑡 = 𝑑𝑣(𝑡) 𝑑𝑡 = 𝑟 𝑡 = 𝑥 𝑡 𝑖+ 𝑦 𝑡 𝑗, we can obtain the velocity vector 𝑣 𝑡 = 𝑟 𝑡 = 𝑥 𝑡 𝑑𝑡 𝑖+ 𝑦 (𝑡) 𝑑𝑡 𝑗+ 𝑐 1 where 𝑐 1 is a constant vector.

Worked examples - cambridge
Given that 𝑟 𝑡 =10 𝑖−12𝑘, find: 𝑟 𝑡 if 𝑟 0 =30 𝑖 −20 𝑗+10𝑘 𝑟(𝑡) if also 𝑟(0) = 0𝑖 + 0𝑗 + 2𝑘 Given 𝑟 𝑡 =−9.8𝑗 with 𝑟(0)=0 and 𝑟 0 =30𝑖+40𝑗, find 𝑟(𝑡).

Worked examples - jacplus
The velocity vector of a particle is given by 𝑟 𝑡 =2𝑖+6𝑡𝑗 for 𝑡≥0. If 𝑟(1)=3𝑖 + 𝑗, find the position vector at time 𝑡. The acceleration vector of a particle is given by 𝑟 𝑡 =6𝑡𝑖, where 𝑡≥0 is the time. Given that 𝑟 2 =6𝑖−3𝑗 and 𝑟(2) = 4𝑖 − 2𝑗, find the position vector at time 𝑡.

Worked examples - jacplus
The acceleration vector of a moving particle is given by cos 𝑡 2 𝑖 − sin 𝑡 2 𝑗 for 0≤𝑡≤4𝜋, where 𝑡 is the time. The initial velocity is 2𝑗 and the initial position is 𝑖−3𝑗. Find the Cartesian equation of the path.

Velocity and Acceleration for motion along a curve – 12D Cambridge

Recap – fill in the blanks
Consider a particle moving along a curve in the plane, with position vector at time 𝑡 given by: We can find the particle’s velocity at time 𝑡 as follows. Velocity Acceleration Velocity is ______________________ Therefore 𝑣(𝑡), the velocity ____________, is given by: The velocity gives _______________________ Acceleration is __________________ Therefore a(t), the acceleration __________________, is given by Speed Distance between two points on the curve Speed is __________________________________. At time t, the speed is denoted by _________________ The _______________ distance between two points on the curve is found using:

Worked Examples Worked Example 15 Worked Example 16 Worked Example 18
The position of an object is 𝑟(𝑡) metres at time 𝑡 seconds, where 𝑟 𝑡 = 𝑒 𝑡 𝑖 𝑒 2𝑡 𝑗, 𝑡≥0. Find at time 𝑡: The velocity The acceleration vector The speed The position vector of a particle at time 𝑡 is given by 𝑟 𝑡 = 2𝑡− 𝑡 2 𝑖+ 𝑡 2 −3𝑡 𝑗+2𝑡𝑘, where 𝑡≥0. Find: The velocity of the particle at time 𝑡 The speed of the particle at time 𝑡 The minimum speed of the particle The position vector of a particle at time 𝑡 is given by 𝑟 𝑡 =2 sin 2𝑡 𝑖+ cos 2𝑡 𝑗+2𝑡 𝑘 The velocity at time 𝑡 The speed of the particle at time 𝑡 The maximum speed The minimum speed

Worked Example - Cambridge
The position of a projectile at time t is give by 𝑟 𝑡 =400𝑡 𝑖+ 500𝑡−5 𝑡 2 𝑗 for 𝑡≥0 where i is a unit vector in a horizontal direction and j is a unit vector vertically up. The projectile is fired from a point on the ground. Find: The time taken to reach the ground again The speed at which the projectile hits the ground The maximum height of the projectile The initial speed of the projectile

The position vectors, at time 𝑡≥0, of particles 𝐴 and 𝐵 are given by 𝑟 𝐴 𝑡 = 𝑡 3 −9𝑡+8 𝑖+ 𝑡 2 𝑗 𝑟 𝐵 𝑡 = 2− 𝑡 2 𝑖+ 3𝑡−2 𝑗 Prove that 𝐴 and 𝐵 collide while travelling at the same speed but at right angles to each other.

Worked examples - Cambridge
A particle moves along a line such that its position at time 𝑡 is given by the vector function: 𝑟 𝑡 = 3𝑡−2 𝑖+ 4𝑡+3 𝑗, 𝑡≥0 How far along the line does the particle travels from 𝑡=1 to 𝑡=3? A particle moves along a curve such that its position vector at time 𝑡 is given by: 𝑟 𝑡 = sin 𝑡 𝑖 sin 2𝑡 𝑗, 𝑡≥ 0 How far along the curve does the particle travel from 𝑡=0 to 𝑡= 𝜋 3 ? (Give your answers to three decimal places.) Find the shortest distance between these two points.

Vector Calculus.

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Vector Calculus.

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Presentation on theme: "Vector Calculus."— Presentation transcript:

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