solving motion problems using parametric and vector valued functions

AP Calculus BC: 9.6 Solving Motion Problems Using Parametric and Vector- Valued Functions – Exam Style questions with Answer- FRQ

For time t ≥ 0, a particle moves in the xy-plane with position (x(t),  y(t)) , and velocity vector \(\left \langle (t-1)^{e^{t^{2}}}, sin \left ( t^{1.25} \right ) \right \rangle.\) At time t = 0 , the position of the particle is (−2, 5) . (a) Find the speed of the particle at time t = 1.2. Find the acceleration vector of the particle at time t = 1.2. (b) Find the total distance traveled by the particle over the time interval 0 ≤ t ≤ 1.2. (c) Find the coordinates of the point at which the particle is farthest to the left for t ≥ 0. Explain why there is no point at which the particle is farthest to the right for t ≥ 0.

\(Speed = \sqrt{(x'(t))^{2}+(y'(t))^{2}}\)

\(\sqrt{\left ( (1.2-1)e^{(1.2)^{2}} \right )^{2}+\left ( sin((1.2)^{1.25}) \right )^{2}}\)

Speed = 1.271                  \(accel: \left \langle (t-1)(2+e^{t2})+(1)(e^{t2}),cos (t^{1.25}\cdot 1.25t^{.25}) \right \rangle\)

of     t = 1.2

accel = < 6.247,   .405>

\(\int_{0}^{1.2}\sqrt{\left ( (t-1)e^{t2} \right )^{2}+\left ( sin(t^{1.25}) \right )^{2}}dt\)

x'(t) is negative on (0, 1) and is positive on (1, 0). x'(1) = 0 Since x'(t) goes from negative to positive at  t = 1 and x'(1) = 0, x(t) has a rel. minimum at  t=1, since x(t) decreases until t=1 and always increases afterwards, x(1) is on abs. min / that is the location of further to the left,   \(-2+\int_{0}^{1}(t-1)e^{t2}dt = x(1)\)

x(1) = -2.604

\(5+\int_{0}^{0}(sin (e^{1.25}))dt = y(1)\)

y (1) = 5.410

(-2.604, 5.410)

There is no point furthest to the right because x'(t) increases towards infinity on +20, meaning there is no abs max of x(t) (sint x(t) will thus be increasing towards ∝  on t > 1). This means the particle will continue to the right for t > 1, creating no furthest right point.

 There are two vectors 〈1, -4〉 and 〈2, k〉 where k is an unknown quantity. (A) Find a value of k such that the vectors are orthogonal. (B) Find a value of k such that the vectors are parallel. (C) If k = 6, find the angle between the two vectors. Round to the nearest tenth of a degree.

(A) If vectors are orthogonal, the angle between them is \(\frac{\pi }{2}\) so that their dot product is defined by \(r_{1}r_{2}\cos \Theta =0\) Then \(r_{1}r_{2}=x_{1}x_{2}+y_{1}y_{2}=\left \langle 1,-4 \right \rangle.\left \langle 2,k \right \rangle=1(2)+(-4)(k)=0\).Then \(-4k=-2\Rightarrow k=\frac{1}{2}\)

(B)If vectors are parallel, the angle between them is 0, where cos(0) = 1. You know that, in the case of parallel vectors,\(r_2=Cr_1\) , where C is a constant. In this case, the horizontal component of \(r_{2}\) differs from \(r_{1}\) by a factor of 2. Multiplying the y-component by the same factor, you get k = -8. You can get the same result by applying the equation \(\cos \Theta =\frac{r_{1}.r_{2}}{\left \| r_{1} \right \|\left \| r_{2} \right \|}=-1\). You find that \(\frac{2-4k}{(\sqrt{17})\sqrt{4+k^{2}}}=-1\). If you multiply through by the denominator and then square both sides, you find that \(4-16k+16k^{2}=17(4+k^{2})\) and, combining terms, you get \(k^{2}+16k+64=0\). You can then factor to find that (k + 8)(k + 8) = 0 or k = -8. (C) Find the angle between two vectors by applying the equation \(\Theta =\cos ^{-1}\frac{r_1.r_2}{\left \| r_{1} \right \|\left \| r_{2} \right \|}.\)Here,\(r_{1}.r_{2}=1(2)-4(6)=-22,\left \| r_{1} \right \|=\sqrt{1^{2}+4^{2}}=\sqrt{17}\)  and \( \left \| r_{2} \right \|=\sqrt{2^{2}+6^{2}}=\sqrt{40}=2\sqrt{10}.\)Substituting, we find that \(\Theta =-\frac{22}{2\sqrt{170}}\approx 147.5^{\circ}\)

 A force of 2 newtons is applied to the right side of an object. A force of 4 newtons is applied from below. A force of 8 newtons is applied to the left side of the object at an angle of \(\frac{\pi }{3}\) above the horizontal. (A) Find the resultant force vector being applied to the object. (B) Find an approximation (to the nearest hundredth) of the magnitude of the total force on the object. (C) What additional force vector would need to be applied to keep the object from moving?

 An object is moving on a path defined by the vector-valued function \(r(t)=\left \langle 2\cos t,2\sin t \right \rangle\).Find the following functions at time \(t=\frac{\pi }{4}\). (A) the velocity vector (B) the acceleration vector (C) the tangent vector (D) the normal vector.

 A curve is defined by parametric equations \( x(t)=t^{3}-2t+1\) and \(y(t)=t^{2}-5t\). (A) Find an equation for the tangent line corresponding to t = 1. (B) For what value(s) of t is the tangent line horizontal? (C) For what value(s) of t is the tangent line vertical?

(A) The slope of the tangent line equals\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}=\frac{2t-5}{3t^{2}-2}-3\) at t=-1 Substituting t = 1 into the parametric equations gives us the coordinates (0, -4). Thus, the equation for the tangent line is y+4=-3(x-0)or,solving fory,y=-3x-4. (B) The tangent line is horizontal if \(\frac{\mathrm{d} y}{\mathrm{d} t}=0\).Thus 2t-5=0 and \(t=\frac{5}{2}\) (C) The tangent line is vertical if \(\frac{\mathrm{d} x}{\mathrm{d} t}=9.\)Thus \(3t^{2}-2=0\) and \(t=\pm \sqrt{\frac{2}{3}}\)

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2: Vector-Valued Functions and Motion in Space

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• 2.1: Vector Valued Functions A vector valued function is a function where the domain is a subset of the real numbers and the range is a vector. There is an equivalence between vector valued functions and parametric equations.
• 2.2: Arc Length in Space For this topic, we will be learning how to calculate the length of a curve in space. The ideas behind this topic are very similar to calculating arc length for a curve in with x and y components, but now, we are considering a third component, z.
• 2.3: Curvature and Normal Vectors of a Curve For a parametrically defined curve we had the definition of arc length. Since vector valued functions are parametrically defined curves in disguise, we have the same definition. We have the added benefit of notation with vector valued functions in that the square root of the sum of the squares of the derivatives is just the magnitude of the velocity vector.
• 2.4: The Unit Tangent and the Unit Normal Vectors The derivative of a vector valued function gives a new vector valued function that is tangent to the defined curve. The analog to the slope of the tangent line is the direction of the tangent line. Since a vector contains a magnitude and a direction, the velocity vector contains more information than we need. We can strip a vector of its magnitude by dividing by its magnitude.
• 2.5: Velocity and Acceleration In single variable calculus the velocity is defined as the derivative of the position function. For vector calculus, we make the same definition.
• 2.6: Tangential and Normal Components of Acceleration This section breaks down acceleration into two components called the tangential and normal components. Similar to how we break down all vectors into \(\hat{\textbf{i}}\), \( \hat{\textbf{j}} \), and \( \hat{\textbf{k}} \) components, we can do the same with acceleration. The addition of these two components will give us the overall acceleration.
• 2.7: Parametric Surfaces We have now seen many kinds of functions. When we talked about parametric curves, we defined them as functions from \(\mathbb{R}\) to \(\mathbb{R}^2\) (plane curves) or \(\mathbb{R}\) to \(\mathbb{R}^3\) (space curves). Because each of these has its domain \(\mathbb{R}\), they are one dimensional (you can only go forward or backward). In this section, we investigate how to parameterize two dimensional surfaces.