Day 4: Motion Along a Curve Vectors

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1 Day 4: Motion Along a Curve Vectors I give my stuents the following list of terms an formulas to know. Parametric Equations, Vectors, an Calculus Terms an Formulas to Know: If a smooth curve C is given by the equations x f t an y g t, then the slope of C at the point (x, y) is given by where 0, an the secon erivative is given by y. = ( ) The erivative also may be interprete as the slope of the tangent line to the curve C, or as the slope of the path of a particle traveling along the curve C, or as the rate of change of y with respect to x. The secon erivative y respect to x. is the rate of change of the slope of the curve C with x ( t) = is the rate at which the x-coorinate is changing with respect to t or the velocity of the particle in the horizontal irection. y ( t) = is the rate at which the y-coorinate is changing with respect to t or the velocity of the particle in the vertical irection The College Boar.

2 x( t), y( t) is the position vector at any time t. x ( t), y ( t) is the velocity vector at any time t. x ( t), y ( t) is the acceleration vector at any time t. velocity vector. b a + + is the spee of the particle or the magnitue (length) of the is the length of the arc (or arc length) of the curve from t = a to t = b or the istance travele by the particle from t = a to t = b. Most textbooks o not contain the types of problems on vectors that are foun on the AP Exam, so I supplement with the examples an worksheets below. Example (no calculator): A particle moves in the xy-plane so that at any time t, the position of the particle is given 4 by x t t 4t, y t t t. + (a) Fin the velocity vector when t =. v( t) = t t t t t t t, = ( + ) ( ) = +, 4 4 8, 4 t v, (b) Fin the acceleration vector when t =. a t a 0, ( ) t t, = + 8, + 4t t t 8, t t Example (no calculator): A particle moves in the xy-plane so that at any time t, t 0, the position of the particle is given by x t t t, y t t t. Fin the magnitue of the velocity vector when t = The College Boar.

3 The magnitue or length of the velocity vector can be foun by using the Pythagorean Theorem, since the horizontal an vertical components make a right triangle, with the vector itself as the hypotenuse. Therefore its length is given by: Magnitue of velocity vector = + For our problem, = t + t t = + an = t t t t =. ( ) + ( ) Magnitue of velocity vector = t + t t t = = = 4. Notice that the formula for the magnitue of the velocity vector is the same as the formula for the spee of the vector, which makes sense since spee is the magnitue of velocity. Example (no calculator): A particle moves in the xy-plane so that x = 4cos t an y = sin t, where 0 t π. The path of the particle intersects the x-axis twice. Write an expression that represents the istance travele by the particle between the two x-intercepts. Do not evaluate. The path of the particle intersects the x-axis at the points where the y-component is equal to zero. Note that sin t = 0 when sin t =. For 0 t π, this will occur when π t = an t = 5π. Since = t t 4cos = 4 sin an = t t sin = cos, 5π the istance travele by the particle is Distance = 4sin t cos t. π ( ) + ( ) 9 00 The College Boar.

4 Day 4 Homework Use your calculator on problems 0 an c only. t. If x = t an y = e, fin.. If a particle moves in the xy-plane so that at any time t > 0, its position vector is ( ), fin its velocity vector at time t =. ln t + 5t, t. A particle moves in the xy-plane so that at any time t, its coorinates are given by 5 4 x = t an y = t t. Fin its acceleration vector at t =. 4. If a particle moves in the xy-plane so that at time t its position vector is π sin t, t, fin the velocity vector at time t = π. 5. A particle moves on the curve y = ln x so that its x-component has erivative + for x t t t 0.At time t = 0, the particle is at the point (, 0). Fin the position of the particle at time t =.. A particle moves in the xy-plane in such a way that its velocity vector is + t, t. If the position vector at t = 0 is 5, 0, fin the position of the particle at t =. 7. A particle moves along the curve xy = 0. If x = an =, what is the value of? 8. The position of a particle moving in the xy-plane is given by the parametric equations x = t t 8t + 5 an y = t t + 9t + 4. For what value(s) of t is the particle at rest? 9. A curve C is efine by the parametric equations x = t an y = t 5t +.Write the equation of the line tangent to the graph of C at the point (8, 4). 0. A particle moves in the xy-plane so that the position of the particle is given by + ( )( ) x t 5 t sin t an y t 8 t cos t. Fin the velocity vector at the time when the particle s horizontal position is x = 5.. The position of a particle at any time t 0 is given by x( t) = t an y( t) = t. (a) Fin the magnitue of the velocity vector at time t = The College Boar.

5 (b) Fin the total istance travele by the particle from t = 0 to t = 5. (c) Fin. Point P x, y = t for t 0. as a function of x. ( ) moves in the xy-plane in such a way that = t + an (a) Fin the coorinates of P in terms of t given that, when t =, x = ln an y = 0. (b) Write an equation expressing y in terms of x. (c) Fin the average rate of change of y with respect to x as t varies from 0 to 4. () Fin the instantaneous rate of change of y with respect to x when t =.. Consier the curve C given by the parametric equations x = cos t an π π y = + sin t, for t. (a) Fin as a function of t. (b) Fin the equation of the tangent line at the point where t = π 4. (c) The curve C intersects the y-axis twice. Approximate the length of the curve between the two y-intercepts. Answers to Day 4 Homework. = = t e t t e te t = = t. t. v t. v t t + 5,, t v, t + 5t 9 so 4. =, 5t, t t = a t 4. v t 4. x y, 0t, t t, so a 0, 4. = π π, cos t t v, so = cos π, π =, π. = 00 The College Boar.

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