Calculus BC ยท Unit 9
Work one component at a time. Speed is the length of the velocity, and rest means both components are zero.
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Differentiate and integrate componentwise
If \(\mathbf{r}(t)=\langle x(t), y(t)\rangle\), then
\(\mathbf{r}'(t)=\langle x'(t), y'(t)\rangle\)
Velocity, acceleration, and speed
Velocity is \(\mathbf{r}'\). Acceleration is \(\mathbf{r}''\). Speed is the length of the velocity vector.
An initial vector picks the constant
\(\mathbf{r}(t)=\mathbf{r}(0)+\int_0^t \mathbf{v}(u)\,du\)
The zero vector is the only rest vector
The object is at rest when both components of velocity are zero at the same \(t\).
One derivative or one antiderivative, component by component.
\(\mathbf{r}(t)=\left\langle t^2+C_1,\ 3t+C_2\right\rangle\)
\(\mathbf{r}(0)=\langle C_1, C_2\rangle=\langle 1, 4\rangle\)
\(\mathbf{r}(t)=\langle t^2+1,\ 3t+4\rangle\)
Answer: \(\langle t^2+1,\ 3t+4\rangle\)
\(\langle 3t^2, \cos t\rangle\)
Answer: \(\langle 3t^2, \cos t\rangle\)
Velocity at \(t=1\) is \(\langle 2, 3\rangle\).
\(\sqrt{4+9}\)
\(\sqrt{13}\)
Answer: \(\sqrt{13}\)
\(\langle 2, 0\rangle\)
Answer: \(\langle 2, 0\rangle\)
\(\sqrt{16+9}\)
\(5\)
Answer: \(5\)
\(\mathbf{v}(t)=\langle 2,\ 6t-1\rangle\)
Answer: \(\langle 2,\ 6t-1\rangle\)
Speed at a point, or a position that uses both components.
\(\mathbf{v}(t)=\langle 2t, 3\rangle\)
At \(t=2\): \(\langle 4, 3\rangle\).
\(\sqrt{16+9}=5\)
Answer: velocity \(\langle 4, 3\rangle\), speed 5
\(\int \cos t\,dt=\sin t\)
\(\int \sin t\,dt=-\cos t\)
\(\mathbf{r}(t)=\langle \sin t+C_1,\ -\cos t+C_2\rangle\)
\(\mathbf{r}(0)=\langle 0,\ -1+C_2\rangle=\langle 0, 1\rangle\)
\(C_2=2\)
\(\mathbf{r}(t)=\langle \sin t,\ -\cos t+2\rangle\)
Answer: \(\langle \sin t,\ 2-\cos t\rangle\)
\(t-1=0\) gives \(t=1\).
At that time the second component is \(2\), not 0.
The components are never zero together.
Answer: no
\(\langle 2, 7\rangle\)
Answer: \(\langle 2, 7\rangle\)
\(\mathbf{v}(t)=\langle 3t^2+1,\ 2t\rangle\)
Answer: \(\langle 3t^2+1,\ 2t\rangle\)
\(\mathbf{r}(t)=\left\langle t^3+t+C_1,\ t^2+C_2\right\rangle\)
\(\mathbf{r}(0)=\langle 0, 3\rangle\) gives \(C_1=0\) and \(C_2=3\).
\(\mathbf{r}(t)=\langle t^3+t,\ t^2+3\rangle\)
Answer: \(\langle t^3+t,\ t^2+3\rangle\)
Displacement is a vector. Distance is a scalar.
Displacement: \(\int_0^2 \langle t-1,\ 2t-2\rangle\,dt\).
\(\left\langle \left[\dfrac{t^2}{2}-t\right]_0^2,\ \left[t^2-2t\right]_0^2\right\rangle\)
\(\langle 0, 0\rangle\)
Speed: \(\sqrt{(t-1)^2+4(t-1)^2}=\sqrt{5}\,|t-1|\).
\(\sqrt{5}\int_0^1 (1-t)\,dt+\sqrt{5}\int_1^2 (t-1)\,dt\)
Each piece equals \(\sqrt{5}/2\).
\(\sqrt{5}\)
Answer: displacement \(\langle 0,0\rangle\), distance \(\sqrt{5}\)
\(t-1=0\) and \(2(t-1)=0\) at the same time.
\(t=1\)
Answer: \(t=1\)
\(\mathbf{v}(t)=\langle 2e^{2t}, 3t^2\rangle\)
\(\mathbf{a}(t)=\langle 4e^{2t}, 6t\rangle\)
At 0: \(\langle 4, 0\rangle\).
Answer: \(\langle 4, 0\rangle\)
\(\mathbf{v}=\langle -3\sin t, 3\cos t\rangle\)
\(|\mathbf{v}|=\sqrt{9\sin^2 t+9\cos^2 t}=3\)
\(x^2+y^2=9\), a circle of radius 3.
Answer: speed 3 on the circle \(x^2+y^2=9\)
Displacement has direction. It is an integral of the velocity vector.
Speed, distance, and arc length are nonnegative scalars.
Answer: A. displacement
\(\mathbf{v}(t)=\langle 2t,\ -2t+4\rangle\)
\(\mathbf{r}(t)=\langle t^2+1,\ -t^2+4t\rangle\)
\(\mathbf{r}(2)=\langle 5, 4\rangle\)
Answer: \(\langle 5, 4\rangle\)
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