Study Defining And Differentiating Vector Valued Functions in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Calculate the speed for r ( t ) = ( sin ( t ) cos ( t ) ) \text{r}(t) = \begin{pmatrix} \text{sin}(t) \ \text{cos}(t) \end{pmatrix} r ( t ) = ( sin ( t ) cos ( t ) ) . Answer: Speed is 1. ∣ v ( t ) ∣ = cos 2 ( t ) + sin 2 ( t ) = 1 |\text{v}(t)| = \sqrt{\cos^2(t) + \sin^2(t)} = 1 ∣ v ( t ) ∣ = cos 2 ( t ) + sin 2 ( t ) = 1 .
Flashcard 2: What does the derivative of a vector-valued function represent? Answer: The tangent vector to the curve at t t t . Points in the direction of motion along the curve.
Flashcard 3: Calculate the unit tangent vector for r ( t ) = ( 3 t 4 t ) \text{r}(t) = \begin{pmatrix} 3t \ 4t \end{pmatrix} r ( t ) = ( 3 t 4 t ) . Answer: T ( t ) = ( 3 5 4 5 ) \text{T}(t) = \begin{pmatrix} \frac{3}{5} \ \frac{4}{5} \end{pmatrix} T ( t ) = ( 5 3 5 4 ) . ∣ r ′ ( t ) ∣ = 9 + 16 = 5 |\text{r}'(t)| = \sqrt{9 + 16} = 5 ∣ r ′ ( t ) ∣ = 9 + 16 = 5 , so T ( t ) = 1 5 ( 3 4 ) \text{T}(t) = \frac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix} T ( t ) = 5 1 ( 3 4 ) .
Flashcard 4: In which dimension is the vector-valued function r ( t ) = ( t t 2 ) \text{r}(t) = \begin{pmatrix} t \ t^2 \end{pmatrix} r ( t ) = ( t t 2 ) ? Answer: 2D. Two components means output lies in 2D space.
Flashcard 5: State the formula for the normal component of acceleration a n ( t ) \text{a}_n(t) a n ( t ) . Answer: a n ( t ) = ∣ r ′ ( t ) × r ′ ′ ( t ) ∣ ∣ r ′ ( t ) ∣ \text{a}_n(t) = \frac{|\text{r}'(t) \times \text{r}''(t)|}{|\text{r}'(t)|} a n ( t ) = ∣ r ′ ( t ) ∣ ∣ r ′ ( t ) × r ′′ ( t ) ∣ . Component of acceleration perpendicular to velocity.
Flashcard 6: State the formula for the arc length of a curve r ( t ) \text{r}(t) r ( t ) from a a a to b b b . Answer: L = ∫ a b ∣ r ′ ( t ) ∣ d t L = \int_a^b |r'(t)| dt L = ∫ a b ∣ r ′ ( t ) ∣ d t . Integrates the magnitude of the velocity vector.
Flashcard 7: Find the velocity for r ( t ) = ( sin ( t ) cos ( t ) ) r(t) = \begin{pmatrix} \sin(t) \\ \cos(t) \end{pmatrix} r ( t ) = ( sin ( t ) cos ( t ) ) . Answer: v ( t ) = ( cos ( t ) − sin ( t ) ) v(t) = \begin{pmatrix} \cos(t) \\ -\sin(t) \end{pmatrix} v ( t ) = ( cos ( t ) − sin ( t ) ) . Differentiate each component: d d t ( sin ( t ) ) = cos ( t ) \frac{d}{dt}(\sin(t)) = \cos(t) d t d ( sin ( t )) = cos ( t ) , d d t ( cos ( t ) ) = − sin ( t ) \frac{d}{dt}(\cos(t)) = -\sin(t) d t d ( cos ( t )) = − sin ( t ) .
Flashcard 8: What is the derivative of a vector-valued function r ( t ) \text{r}(t) r ( t ) ? Answer: r ′ ( t ) = ( x ′ ( t ) y ′ ( t ) ) \text{r}'(t) = \begin{pmatrix} x'(t) \\ y'(t) \end{pmatrix} r ′ ( t ) = ( x ′ ( t ) y ′ ( t ) ) . Differentiate each component separately.
Flashcard 9: What is the geometric interpretation of speed in vector-valued functions? Answer: Magnitude of the velocity vector. Speed is scalar, velocity includes direction.
Flashcard 10: What is the formula for the torsion τ ( t ) \text{τ}(t) τ ( t ) of a curve? Answer: τ ( t ) = − ( r ′ ( t ) × r ′ ′ ( t ) ) ∙ r ′ ′ ′ ( t ) ∣ r ′ ( t ) × r ′ ′ ( t ) ∣ 2 \text{τ}(t) = -\frac{(\text{r}'(t) \times \text{r}''(t)) \bullet \text{r}'''(t)}{|\text{r}'(t) \times \text{r}''(t)|^2} τ ( t ) = − ∣ r ′ ( t ) × r ′′ ( t ) ∣ 2 ( r ′ ( t ) × r ′′ ( t )) ∙ r ′′′ ( t ) . Measures how much the curve twists out of its plane.
Flashcard 11: How is the binormal vector B ( t ) \text{B}(t) B ( t ) defined? Answer: B ( t ) = T ( t ) × N ( t ) \text{B}(t) = \text{T}(t) \times \text{N}(t) B ( t ) = T ( t ) × N ( t ) . Cross product of tangent and normal vectors.
Flashcard 12: How do you represent a vector-valued function in 2D? Answer: r ( t ) = ( x ( t ) y ( t ) ) \text{r}(t) = \begin{pmatrix} x(t) \ y(t) \end{pmatrix} r ( t ) = ( x ( t ) y ( t ) ) . Components x ( t ) x(t) x ( t ) and y ( t ) y(t) y ( t ) form a 2D vector output.
Flashcard 13: What is the speed of the particle moving along r ( t ) \text{r}(t) r ( t ) ? Answer: Speed is ∣ v ( t ) ∣ = ∣ r ′ ( t ) ∣ |\text{v}(t)| = |\text{r}'(t)| ∣ v ( t ) ∣ = ∣ r ′ ( t ) ∣ . Magnitude gives distance traveled per unit time.
Flashcard 14: What is the relationship between velocity and speed? Answer: Speed is the magnitude of velocity. Speed removes directional information from velocity.
Flashcard 15: Find the velocity for r ( t ) = ( e t ln ( t ) ) \text{r}(t) = \begin{pmatrix} e^t \\ \text{ln}(t) \end{pmatrix} r ( t ) = ( e t ln ( t ) ) . Answer: v ( t ) = ( e t 1 t ) \text{v}(t) = \begin{pmatrix} e^t \\ \frac{1}{t} \end{pmatrix} v ( t ) = ( e t t 1 ) . Differentiate: d d t ( e t ) = e t \frac{d}{dt}(e^t) = e^t d t d ( e t ) = e t , d d t ( ln ( t ) ) = 1 t \frac{d}{dt}(\ln(t)) = \frac{1}{t} d t d ( ln ( t )) = t 1 .
Flashcard 16: Find the derivative: r ( t ) = ( t 2 1 t ) \text{r}(t) = \begin{pmatrix} t^2 \\ \frac{1}{t} \end{pmatrix} r ( t ) = ( t 2 t 1 ) Answer: r ′ ( t ) = ( 2 t − 1 t 2 ) \text{r}'(t) = \begin{pmatrix} 2t \\ -\frac{1}{t^2} \end{pmatrix} r ′ ( t ) = ( 2 t − t 2 1 ) . Differentiate: d d t ( t 2 ) = 2 t \frac{d}{dt}(t^2) = 2t d t d ( t 2 ) = 2 t and d d t ( 1 t ) = − 1 t 2 \frac{d}{dt}(\frac{1}{t}) = -\frac{1}{t^2} d t d ( t 1 ) = − t 2 1
Flashcard 17: State the definition of the unit normal vector N ( t ) \text{N}(t) N ( t ) . Answer: A vector perpendicular to the unit tangent vector. Points toward center of curvature, perpendicular to tangent.
Flashcard 18: How do you determine the normal vector N ( t ) \text{N}(t) N ( t ) ? Answer: N ( t ) = T ′ ( t ) ∣ T ′ ( t ) ∣ \text{N}(t) = \frac{\text{T}'(t)}{|\text{T}'(t)|} N ( t ) = ∣ T ′ ( t ) ∣ T ′ ( t ) . Normalized derivative of unit tangent vector.
Flashcard 19: Identify the acceleration vector of r ( t ) \text{r}(t) r ( t ) . Answer: a ( t ) = r ′ ′ ( t ) \text{a}(t) = \text{r}''(t) a ( t ) = r ′′ ( t ) . Rate of change of velocity vector.
Flashcard 20: Find the velocity for r ( t ) = ( sin ( t ) cos ( t ) ) \text{r}(t) = \begin{pmatrix} \text{sin}(t) \ \text{cos}(t) \end{pmatrix} r ( t ) = ( sin ( t ) cos ( t ) ) . Answer: v ( t ) = ( cos ( t ) − sin ( t ) ) \text{v}(t) = \begin{pmatrix} \text{cos}(t) \ -\text{sin}(t) \end{pmatrix} v ( t ) = ( cos ( t ) − sin ( t ) ) . Differentiate each component: d d t ( sin ( t ) ) = cos ( t ) \frac{d}{dt}(\sin(t)) = \cos(t) d t d ( sin ( t )) = cos ( t ) , d d t ( cos ( t ) ) = − sin ( t ) \frac{d}{dt}(\cos(t)) = -\sin(t) d t d ( cos ( t )) = − sin ( t ) .
Flashcard 21: State the formula for the velocity vector of r ( t ) \text{r}(t) r ( t ) . Answer: v ( t ) = r ′ ( t ) \text{v}(t) = \text{r}'(t) v ( t ) = r ′ ( t ) . Rate of change of position vector.
Flashcard 22: What is the derivative of a vector-valued function r ( t ) \text{r}(t) r ( t ) ? Answer: r ′ ( t ) = ( x ′ ( t ) y ′ ( t ) ) \text{r}'(t) = \begin{pmatrix} x'(t) \ y'(t) \end{pmatrix} r ′ ( t ) = ( x ′ ( t ) y ′ ( t ) ) . Differentiate each component separately.
Flashcard 23: Find the velocity for r ( t ) = ( e t ln ( t ) ) \text{r}(t) = \begin{pmatrix} e^t \\ \ln(t) \end{pmatrix} r ( t ) = ( e t ln ( t ) ) . Answer: v ( t ) = ( e t 1 t ) \text{v}(t) = \begin{pmatrix} e^t \\ \frac{1}{t} \end{pmatrix} v ( t ) = ( e t t 1 ) . Differentiate: d d t ( e t ) = e t \frac{d}{dt}(e^t) = e^t d t d ( e t ) = e t , d d t ( ln ( t ) ) = 1 t \frac{d}{dt}(\ln(t)) = \frac{1}{t} d t d ( ln ( t )) = t 1 .
Flashcard 24: What is the curvature formula for a vector-valued function r ( t ) \text{r}(t) r ( t ) ? Answer: k ( t ) = ∣ r ′ ( t ) × r ′ ′ ( t ) ∣ ∣ r ′ ( t ) ∣ 3 \text{k}(t) = \frac{|\text{r}'(t) \times \text{r}''(t)|}{|\text{r}'(t)|^3} k ( t ) = ∣ r ′ ( t ) ∣ 3 ∣ r ′ ( t ) × r ′′ ( t ) ∣ . Measures how sharply the curve bends.
Flashcard 25: How do you express the tangent vector T ( t ) \text{T}(t) T ( t ) using derivatives? Answer: T ( t ) = r ′ ( t ) ∣ r ′ ( t ) ∣ \text{T}(t) = \frac{\text{r}'(t)}{|\text{r}'(t)|} T ( t ) = ∣ r ′ ( t ) ∣ r ′ ( t ) . Unit vector in direction of velocity.
Flashcard 26: How do you find the position vector from velocity v ( t ) \text{v}(t) v ( t ) ? Answer: Integrate: \text{r}(t) = \text{∫} \text{v}(t) \text{dt} + \text{C} . Antiderivative of velocity plus initial condition.
Flashcard 27: Identify the unit tangent vector of r ( t ) \text{r}(t) r ( t ) . Answer: T ( t ) = r ′ ( t ) ∣ r ′ ( t ) ∣ \text{T}(t) = \frac{\text{r}'(t)}{|\text{r}'(t)|} T ( t ) = ∣ r ′ ( t ) ∣ r ′ ( t ) . Normalizes the velocity vector to unit length.
Flashcard 28: How do you determine the normal vector N ( t ) \text{N}(t) N ( t ) ? Answer: N ( t ) = T ′ ( t ) ∣ T ′ ( t ) ∣ \text{N}(t) = \frac{\text{T}'(t)}{|\text{T}'(t)|} N ( t ) = ∣ T ′ ( t ) ∣ T ′ ( t ) . Normalized derivative of unit tangent vector.
Flashcard 29: How do you find the position vector from velocity v ( t ) v(t) v ( t ) ? Answer: Integrate: r ( t ) = ∫ v ( t ) d t + C r(t) = \int v(t) dt + C r ( t ) = ∫ v ( t ) d t + C . Antiderivative of velocity plus initial condition.
Flashcard 30: Calculate the acceleration for r ( t ) = ( t 3 3 t 2 ) \text{r}(t) = \begin{pmatrix} t^3 \ 3t^2 \end{pmatrix} r ( t ) = ( t 3 3 t 2 ) . Answer: a ( t ) = ( 6 t 6 ) \text{a}(t) = \begin{pmatrix} 6t \ 6 \end{pmatrix} a ( t ) = ( 6 t 6 ) . Second derivative: d 2 d t 2 ( t 3 ) = 6 t \frac{d^2}{dt^2}(t^3) = 6t d t 2 d 2 ( t 3 ) = 6 t , d 2 d t 2 ( 3 t 2 ) = 6 \frac{d^2}{dt^2}(3t^2) = 6 d t 2 d 2 ( 3 t 2 ) = 6 .
Flashcard 31: What is the curvature formula for a vector-valued function r ( t ) \text{r}(t) r ( t ) ? Answer: k ( t ) = ∣ r ′ ( t ) × r ′ ′ ( t ) ∣ ∣ r ′ ( t ) ∣ 3 \text{k}(t) = \frac{|\text{r}'(t) \times \text{r}''(t)|}{|\text{r}'(t)|^3} k ( t ) = ∣ r ′ ( t ) ∣ 3 ∣ r ′ ( t ) × r ′′ ( t ) ∣ . Measures how sharply the curve bends.
Flashcard 32: How do you express the tangent vector T ( t ) \text{T}(t) T ( t ) using derivatives? Answer: T ( t ) = r ′ ( t ) ∣ r ′ ( t ) ∣ \text{T}(t) = \frac{\text{r}'(t)}{|\text{r}'(t)|} T ( t ) = ∣ r ′ ( t ) ∣ r ′ ( t ) . Unit vector in direction of velocity.
Flashcard 33: Identify the acceleration vector of r ( t ) \text{r}(t) r ( t ) . Answer: a ( t ) = r ′ ′ ( t ) \text{a}(t) = \text{r}''(t) a ( t ) = r ′′ ( t ) . Rate of change of velocity vector.
Flashcard 34: Identify the velocity vector for r ( t ) = ( 5 t t 2 t ) \text{r}(t) = \begin{pmatrix} 5t \ t^2 \ t \end{pmatrix} r ( t ) = ( 5 t t 2 t ) . Answer: v ( t ) = ( 5 2 t 1 ) \text{v}(t) = \begin{pmatrix} 5 \ 2t \ 1 \end{pmatrix} v ( t ) = ( 5 2 t 1 ) . Differentiate each component: constants and powers.
Flashcard 35: Calculate the speed for r ( t ) = ( sin ( t ) cos ( t ) ) \text{r}(t) = \begin{pmatrix} \sin(t) \\ \cos(t) \end{pmatrix} r ( t ) = ( sin ( t ) cos ( t ) ) Answer: Speed is 1. ∣ v ( t ) ∣ = cos 2 ( t ) + sin 2 ( t ) = 1 |\text{v}(t)| = \sqrt{\cos^2(t) + \sin^2(t)} = 1 ∣ v ( t ) ∣ = cos 2 ( t ) + sin 2 ( t ) = 1
Flashcard 36: Determine the derivative for r ( t ) = ( 2 t e t ) \text{r}(t) = \begin{pmatrix} 2t \ e^t \end{pmatrix} r ( t ) = ( 2 t e t ) . Answer: r ′ ( t ) = ( 2 e t ) \text{r}'(t) = \begin{pmatrix} 2 \ e^t \end{pmatrix} r ′ ( t ) = ( 2 e t ) . Differentiate: d d t ( 2 t ) = 2 \frac{d}{dt}(2t) = 2 d t d ( 2 t ) = 2 , d d t ( e t ) = e t \frac{d}{dt}(e^t) = e^t d t d ( e t ) = e t .
Flashcard 37: What is the parametric form of a circle with radius R R R ? Answer: r ( t ) = ( R cos ( t ) R sin ( t ) ) r(t) = \begin{pmatrix} R \cos(t) \ R \sin(t) \end{pmatrix} r ( t ) = ( R cos ( t ) R sin ( t ) ) . Standard parametrization using trigonometric functions.
Flashcard 38: What is the speed of the particle moving along r ( t ) \text{r}(t) r ( t ) ? Answer: Speed is ∣ v ( t ) ∣ = ∣ r ′ ( t ) ∣ |\text{v}(t)| = |\text{r}'(t)| ∣ v ( t ) ∣ = ∣ r ′ ( t ) ∣ . Magnitude gives distance traveled per unit time.
Flashcard 39: What is the formula for the torsion τ ( t ) \text{τ}(t) τ ( t ) of a curve? Answer: τ ( t ) = − ( r ′ ( t ) × r ′ ′ ( t ) ) ∙ r ′ ′ ′ ( t ) ∣ r ′ ( t ) × r ′ ′ ( t ) ∣ 2 \text{τ}(t) = -\frac{(\text{r}'(t) \times \text{r}''(t)) \bullet \text{r}'''(t)}{|\text{r}'(t) \times \text{r}''(t)|^2} τ ( t ) = − ∣ r ′ ( t ) × r ′′ ( t ) ∣ 2 ( r ′ ( t ) × r ′′ ( t )) ∙ r ′′′ ( t ) . Measures how much the curve twists out of its plane.
Flashcard 40: State the definition of the unit normal vector N ( t ) \text{N}(t) N ( t ) . Answer: A vector perpendicular to the unit tangent vector. Points toward center of curvature, perpendicular to tangent.
Flashcard 41: Determine the derivative for r ( t ) = ( 2 t e t ) \text{r}(t) = \begin{pmatrix} 2t \ \text{e}^t \end{pmatrix} r ( t ) = ( 2 t e t ) . Answer: r ′ ( t ) = ( 2 e t ) \text{r}'(t) = \begin{pmatrix} 2 \ \text{e}^t \end{pmatrix} r ′ ( t ) = ( 2 e t ) . Differentiate: d d t ( 2 t ) = 2 \frac{d}{dt}(2t) = 2 d t d ( 2 t ) = 2 , d d t ( e t ) = e t \frac{d}{dt}(e^t) = e^t d t d ( e t ) = e t .
Flashcard 42: What is a vector-valued function? Answer: A function with vector outputs, mapping from R \text{R} R to R n \text{R}^n R n . Each input maps to a vector with multiple components.
Flashcard 43: What is the integral of a vector-valued function r ( t ) \text{r}(t) r ( t ) ? Answer: R ( t ) = ( X ( t ) Y ( t ) ) + C \text{R}(t) = \begin{pmatrix} \text{X}(t) \ \text{Y}(t) \end{pmatrix} + \text{C} R ( t ) = ( X ( t ) Y ( t ) ) + C . Integrate each component and add constant vector.
Flashcard 44: State the formula for the arc length of a curve r ( t ) r(t) r ( t ) from a a a to b b b . Answer: L = ∫ a b ∣ r ′ ( t ) ∣ d t L = ∫_{a}^{b} |r'(t)| dt L = ∫ a b ∣ r ′ ( t ) ∣ d t . Integrates the magnitude of the velocity vector.
Flashcard 45: Identify the velocity vector for r ( t ) = ( 5 t t 2 t ) \text{r}(t) = \begin{pmatrix} 5t \ t^2 \ t \end{pmatrix} r ( t ) = ( 5 t t 2 t ) . Answer: v ( t ) = ( 5 2 t 1 ) \text{v}(t) = \begin{pmatrix} 5 \ 2t \ 1 \end{pmatrix} v ( t ) = ( 5 2 t 1 ) . Differentiate each component: constants and powers.
Flashcard 46: Identify the unit tangent vector of r ( t ) \text{r}(t) r ( t ) . Answer: T ( t ) = r ′ ( t ) ∣ r ′ ( t ) ∣ \text{T}(t) = \frac{\text{r}'(t)}{|\text{r}'(t)|} T ( t ) = ∣ r ′ ( t ) ∣ r ′ ( t ) . Normalizes the velocity vector to unit length.
Flashcard 47: What is the geometric interpretation of speed in vector-valued functions? Answer: Magnitude of the velocity vector. Speed is scalar, velocity includes direction.
Flashcard 48: What is the relationship between velocity and speed? Answer: Speed is the magnitude of velocity. Speed removes directional information from velocity.
Flashcard 49: How is the binormal vector B ( t ) \text{B}(t) B ( t ) defined? Answer: B ( t ) = T ( t ) × N ( t ) \text{B}(t) = \text{T}(t) \times \text{N}(t) B ( t ) = T ( t ) × N ( t ) . Cross product of tangent and normal vectors.
Flashcard 50: Calculate the unit tangent vector for r ( t ) = ( 3 t 4 t ) \text{r}(t) = \begin{pmatrix} 3t \ 4t \end{pmatrix} r ( t ) = ( 3 t 4 t ) Answer: T ( t ) = ( 3 5 4 5 ) \text{T}(t) = \begin{pmatrix} \frac{3}{5} \\ \frac{4}{5} \end{pmatrix} T ( t ) = ( 5 3 5 4 ) . ∣ r ′ ( t ) ∣ = 9 + 16 = 5 |\text{r}'(t)| = \sqrt{9 + 16} = 5 ∣ r ′ ( t ) ∣ = 9 + 16 = 5 , so T ( t ) = 1 5 ( 3 4 ) \text{T}(t) = \frac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix} T ( t ) = 5 1 ( 3 4 )
Flashcard 51: What is the integral of a vector-valued function r ( t ) \text{r}(t) r ( t ) ? Answer: R ( t ) = ( X ( t ) Y ( t ) ) + C \text{R}(t) = \begin{pmatrix} \text{X}(t) \\ \text{Y}(t) \end{pmatrix} + \text{C} R ( t ) = ( X ( t ) Y ( t ) ) + C . Integrate each component and add constant vector.
Flashcard 52: In which dimension is the vector-valued function r ( t ) = ( t t 2 ) \text{r}(t) = \begin{pmatrix} t \ t^2 \end{pmatrix} r ( t ) = ( t t 2 ) ? Answer: 2D. Two components means output lies in 2D space.
Flashcard 53: What is the parametric form of a circle with radius R R R ? Answer: $$ Standard parametrization using trigonometric functions.
Flashcard 54: How do you represent a vector-valued function in 2D? Answer: r ( t ) = ( x ( t ) y ( t ) ) \text{r}(t) = \begin{pmatrix} x(t) \ y(t) \end{pmatrix} r ( t ) = ( x ( t ) y ( t ) ) . Components x ( t ) x(t) x ( t ) and y ( t ) y(t) y ( t ) form a 2D vector output.
Flashcard 55: What is a vector-valued function? Answer: A function with vector outputs, mapping from R \text{R} R to R n \text{R}^n R n . Each input maps to a vector with multiple components.
Flashcard 56: Find the derivative: r ( t ) = ( t 2 1 t ) \text{r}(t) = \begin{pmatrix} t^2 \\ \frac{1}{t} \end{pmatrix} r ( t ) = ( t 2 t 1 ) Answer: r ′ ( t ) = ( 2 t − 1 t 2 ) \text{r}'(t) = \begin{pmatrix} 2t \\ -\frac{1}{t^2} \end{pmatrix} r ′ ( t ) = ( 2 t − t 2 1 ) . Differentiate: d d t ( t 2 ) = 2 t \frac{d}{dt}(t^2) = 2t d t d ( t 2 ) = 2 t and d d t ( 1 t ) = − 1 t 2 \frac{d}{dt}(\frac{1}{t}) = -\frac{1}{t^2} d t d ( t 1 ) = − t 2 1
Flashcard 57: State the formula for the normal component of acceleration a n ( t ) \text{a}_n(t) a n ( t ) . Answer: a n ( t ) = ∣ r ′ ( t ) × r ′ ′ ( t ) ∣ ∣ r ′ ( t ) ∣ \text{a}_n(t) = \frac{|\text{r}'(t) \times \text{r}''(t)|}{|\text{r}'(t)|} a n ( t ) = ∣ r ′ ( t ) ∣ ∣ r ′ ( t ) × r ′′ ( t ) ∣ . Component of acceleration perpendicular to velocity.
Flashcard 58: State the formula for the velocity vector of r ( t ) \text{r}(t) r ( t ) . Answer: v ( t ) = r ′ ( t ) \text{v}(t) = \text{r}'(t) v ( t ) = r ′ ( t ) . Rate of change of position vector.
Flashcard 59: What does the derivative of a vector-valued function represent? Answer: The tangent vector to the curve at t t t . Points in the direction of motion along the curve.