Find the unit vector in the direction of v ⃗ = ( 3 , − 4 , 0 ) \vec{v} = (3, -4, 0) v = ( 3 , − 4 , 0 ) .
Solution: v ^ = v ⃗ ∥ v ⃗ ∥ = ( 3 5 , − 4 5 , 0 ) \hat{v} = \frac{\vec{v}}{\|\vec{v}\|} = (\frac{3}{5}, -\frac{4}{5}, 0) v ^ = ∥ v ∥ v = ( 5 3 , − 5 4 , 0 )
Determine the angle between the vectors u ⃗ = ( 1 , 2 , − 1 ) \vec{u} = (1, 2, -1) u = ( 1 , 2 , − 1 ) and v ⃗ = ( 2 , 0 , 3 ) \vec{v} = (2, 0, 3) v = ( 2 , 0 , 3 ) .
Solution: cos θ = u ⃗ ⋅ v ⃗ ∥ u ⃗ ∥ ∥ v ⃗ ∥ = 5 6 13 \cos{\theta} = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|} = \frac{5}{\sqrt{6} \sqrt{13}} cos θ = ∥ u ∥∥ v ∥ u ⋅ v = 6 13 5 , so θ ≈ 49.9 ° \theta \approx 49.9° θ ≈ 49.9°
Find the area of the parallelogram formed by the vectors a ⃗ = ( 1 , 2 ) \vec{a} = (1, 2) a = ( 1 , 2 ) and b ⃗ = ( − 3 , 1 ) \vec{b} = (-3, 1) b = ( − 3 , 1 ) .
Solution: Area = ∥ a ⃗ × b ⃗ ∥ = ∥ ( − 2 , − 7 ) ∥ = 53 = \|\vec{a} \times \vec{b}\| = \|(-2, -7)\| = \sqrt{53} = ∥ a × b ∥ = ∥ ( − 2 , − 7 ) ∥ = 53
Given the vector-valued function r ⃗ ( t ) = ( t 2 , sin t , e t ) \vec{r}(t) = (t^2, \sin{t}, e^t) r ( t ) = ( t 2 , sin t , e t ) , find r ⃗ ′ ( t ) \vec{r}'(t) r ′ ( t ) and r ⃗ ′ ′ ( t ) \vec{r}''(t) r ′′ ( t ) .
Solution: r ⃗ ′ ( t ) = ( 2 t , cos t , e t ) \vec{r}'(t) = (2t, \cos{t}, e^t) r ′ ( t ) = ( 2 t , cos t , e t ) and r ⃗ ′ ′ ( t ) = ( 2 , − sin t , e t ) \vec{r}''(t) = (2, -\sin{t}, e^t) r ′′ ( t ) = ( 2 , − sin t , e t )
Evaluate the definite integral ∫ 0 1 ( 3 t 2 , 2 t , 1 ) d t \int_0^1 (3t^2, 2t, 1) dt ∫ 0 1 ( 3 t 2 , 2 t , 1 ) d t .
Solution: ∫ 0 1 ( 3 t 2 , 2 t , 1 ) d t = ( 1 , 1 , 1 ) \int_0^1 (3t^2, 2t, 1) dt = (1, 1, 1) ∫ 0 1 ( 3 t 2 , 2 t , 1 ) d t = ( 1 , 1 , 1 )
A particle moves along the curve r ⃗ ( t ) = ( 2 cos t , 2 sin t , t ) \vec{r}(t) = (2\cos{t}, 2\sin{t}, t) r ( t ) = ( 2 cos t , 2 sin t , t ) . Find its speed at t = π 4 t = \frac{\pi}{4} t = 4 π .
Solution: Speed = ∥ r ⃗ ′ ( t ) ∥ = ( − 2 sin t ) 2 + ( 2 cos t ) 2 + 1 2 = 5 = \|\vec{r}'(t)\| = \sqrt{(-2\sin{t})^2 + (2\cos{t})^2 + 1^2} = \sqrt{5} = ∥ r ′ ( t ) ∥ = ( − 2 sin t ) 2 + ( 2 cos t ) 2 + 1 2 = 5 at t = π 4 t = \frac{\pi}{4} t = 4 π
Find the work done by the force F ⃗ ( x , y ) = ( x 2 , y ) \vec{F}(x, y) = (x^2, y) F ( x , y ) = ( x 2 , y ) along the path r ⃗ ( t ) = ( t , t 2 ) \vec{r}(t) = (t, t^2) r ( t ) = ( t , t 2 ) from t = 0 t = 0 t = 0 to t = 1 t = 1 t = 1 .
Solution: W = ∫ C F ⃗ ⋅ d r ⃗ = ∫ 0 1 ( t 2 , t 2 ) ⋅ ( 1 , 2 t ) d t = ∫ 0 1 ( t 2 + 2 t 3 ) d t = 7 12 W = \int_C \vec{F} \cdot d\vec{r} = \int_0^1 (t^2, t^2) \cdot (1, 2t) dt = \int_0^1 (t^2 + 2t^3) dt = \frac{7}{12} W = ∫ C F ⋅ d r = ∫ 0 1 ( t 2 , t 2 ) ⋅ ( 1 , 2 t ) d t = ∫ 0 1 ( t 2 + 2 t 3 ) d t = 12 7
Determine the curvature of the helix r ⃗ ( t ) = ( cos t , sin t , t ) \vec{r}(t) = (\cos{t}, \sin{t}, t) r ( t ) = ( cos t , sin t , t ) at t = π 2 t = \frac{\pi}{2} t = 2 π .
Solution: κ ( t ) = ∥ r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) ∥ ∥ r ⃗ ′ ( t ) ∥ 3 = 1 2 \kappa(t) = \frac{\|\vec{r}'(t) \times \vec{r}''(t)\|}{\|\vec{r}'(t)\|^3} = \frac{1}{\sqrt{2}} κ ( t ) = ∥ r ′ ( t ) ∥ 3 ∥ r ′ ( t ) × r ′′ ( t ) ∥ = 2 1 at t = π 2 t = \frac{\pi}{2} t = 2 π