Chapter 9: Parametric Equations, Polar Coordinates, and Vector–Valued Functions# In this chapter we expand our toolbox by studying alternative representations of curves and motion. Instead of expressing functions solely as y = f ( x ) y=f(x) y = f ( x )
9.1 Defining and Differentiating Parametric Equations# Parametric equations express both x x x y y y t t t
x = f ( t ) , y = g ( t ) . x = f(t), \quad y = g(t). x = f ( t ) , y = g ( t ) . To find the slope of the tangent line, we compute
d y d x = d y / d t d x / d t , \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, d x d y = d x / d t d y / d t , provided d x / d t ≠ 0 dx/dt \neq 0 d x / d t = 0
Self–Check 1# Problem:
x = t 2 + 1 , y = t 3 − t , x = t^2 + 1, \quad y = t^3 - t, x = t 2 + 1 , y = t 3 − t , find d y d x \frac{dy}{dx} d x d y t t t
Solution:
Differentiate:d x d t = 2 t , \frac{dx}{dt} = 2t, d t d x = 2 t , d y d t = 3 t 2 − 1. \frac{dy}{dt} = 3t^2 - 1. d t d y = 3 t 2 − 1. Compute the derivative: d y d x = 3 t 2 − 1 2 t . \frac{dy}{dx} = \frac{3t^2 - 1}{2t}. d x d y = 2 t 3 t 2 − 1 . 9.2 Second Derivatives of Parametric Equations# The second derivative, d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y d y d x \frac{dy}{dx} d x d y t t t d x / d t dx/dt d x / d t
d 2 y d x 2 = d d t ( d y d x ) ÷ d x d t . \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}. d x 2 d 2 y = d t d ( d x d y ) ÷ d t d x . Self–Check 2# Problem:
x = t 2 , y = t 3 , x = t^2, \quad y = t^3, x = t 2 , y = t 3 , find d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y t t t
Solution:
First derivative:d x d t = 2 t , \frac{dx}{dt} = 2t, d t d x = 2 t , d y d t = 3 t 2 , \frac{dy}{dt} = 3t^2, d t d y = 3 t 2 , So, d y d x = 3 t 2 2 t = 3 t 2 . \frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}. d x d y = 2 t 3 t 2 = 2 3 t . Differentiate d y d x \frac{dy}{dx} d x d y t t t d d t ( 3 t 2 ) = 3 2 . \frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}. d t d ( 2 3 t ) = 2 3 . Divide by d x / d t dx/dt d x / d t d 2 y d x 2 = 3 2 2 t = 3 4 t . \frac{d^2y}{dx^2} = \frac{\frac{3}{2}}{2t} = \frac{3}{4t}. d x 2 d 2 y = 2 t 2 3 = 4 t 3 . 9.3 Finding Arc Lengths of Curves Given by Parametric Equations# The arc length of a curve defined by parametric equations from t = a t=a t = a t = b t=b t = b
L = ∫ a b ( d x d t ) 2 + ( d y d t ) 2 d t . L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt. L = ∫ a b ( d t d x ) 2 + ( d t d y ) 2 d t . This formula comes from approximating the curve by small line segments and taking the limit as the segment length approaches zero.
Self–Check 3# Problem:
x = 2 t , y = t 2 , x = 2t, \quad y = t^2, x = 2 t , y = t 2 , for t t t [ 0 , 3 ] [0,3] [ 0 , 3 ]
Solution:
Compute derivatives:d x d t = 2 , \frac{dx}{dt} = 2, d t d x = 2 , d y d t = 2 t . \frac{dy}{dt} = 2t. d t d y = 2 t . Set up the integral: L = ∫ 0 3 ( 2 ) 2 + ( 2 t ) 2 d t = ∫ 0 3 4 + 4 t 2 d t . L = \int_0^3 \sqrt{(2)^2 + (2t)^2}\,dt = \int_0^3 \sqrt{4 + 4t^2}\,dt. L = ∫ 0 3 ( 2 ) 2 + ( 2 t ) 2 d t = ∫ 0 3 4 + 4 t 2 d t . Factor out 4: L = ∫ 0 3 2 1 + t 2 d t = 2 ∫ 0 3 1 + t 2 d t . L = \int_0^3 2\sqrt{1+t^2}\,dt = 2\int_0^3 \sqrt{1+t^2}\,dt. L = ∫ 0 3 2 1 + t 2 d t = 2 ∫ 0 3 1 + t 2 d t . (Leave the answer in integral form or evaluate numerically as desired.) 9.4 Defining and Differentiating Vector–Valued Functions# Vector–valued functions express curves in space by specifying each coordinate as a function of a parameter:
r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) , z ( t ) ⟩ . \vec{r}(t) = \langle x(t),\, y(t),\, z(t) \rangle. r ( t ) = ⟨ x ( t ) , y ( t ) , z ( t )⟩ . The derivative, r ⃗ ′ ( t ) \vec{r}'\,(t) r ′ ( t )
r ⃗ ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t ) , z ′ ( t ) ⟩ . \vec{r}'\,(t) = \langle x'(t),\, y'(t),\, z'(t) \rangle. r ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t ) , z ′ ( t )⟩ . Self–Check 4# Problem:
r ⃗ ( t ) = ⟨ cos t , sin t , t ⟩ , \vec{r}(t) = \langle \cos t,\, \sin t,\, t \rangle, r ( t ) = ⟨ cos t , sin t , t ⟩ , find r ⃗ ′ ( t ) \vec{r}'\,(t) r ′ ( t )
Solution:
Differentiate each component:d d t ( cos t ) = − sin t , \frac{d}{dt}(\cos t) = -\sin t, d t d ( cos t ) = − sin t , d d t ( sin t ) = cos t , \frac{d}{dt}(\sin t) = \cos t, d t d ( sin t ) = cos t , d d t ( t ) = 1. \frac{d}{dt}(t) = 1. d t d ( t ) = 1. So, r ⃗ ′ ( t ) = ⟨ − sin t , cos t , 1 ⟩ . \vec{r}'\,(t) = \langle -\sin t,\, \cos t,\, 1 \rangle. r ′ ( t ) = ⟨ − sin t , cos t , 1 ⟩ . 9.5 Integrating Vector–Valued Functions# Integrating a vector–valued function allows us to determine the position function from a velocity function. If
v ⃗ ( t ) = r ⃗ ′ ( t ) , \vec{v}(t) = \vec{r}'\,(t), v ( t ) = r ′ ( t ) , then the position function is given by:
r ⃗ ( t ) = ∫ v ⃗ ( t ) d t . \vec{r}(t) = \int \vec{v}(t)\,dt. r ( t ) = ∫ v ( t ) d t . This process is done component–wise.
Self–Check 5# Problem:
v ⃗ ( t ) = ⟨ 2 , 3 t , 4 e t ⟩ , \vec{v}(t) = \langle 2,\, 3t,\, 4e^t \rangle, v ( t ) = ⟨ 2 , 3 t , 4 e t ⟩ , and the initial condition r ⃗ ( 0 ) = ⟨ 0 , 1 , 2 ⟩ \vec{r}(0)= \langle 0,\, 1,\, 2 \rangle r ( 0 ) = ⟨ 0 , 1 , 2 ⟩
Solution:
Integrate each component:For the first component: ∫ 2 d t = 2 t + C 1 ; C 1 = 0 (since x ( 0 ) = 0 ) . \int 2\,dt = 2t + C_1; \quad C_1=0 \text{ (since } x(0)=0 \text{)}. ∫ 2 d t = 2 t + C 1 ; C 1 = 0 (since x ( 0 ) = 0 ) . Second component: ∫ 3 t d t = 3 t 2 2 + C 2 ; C 2 = 1 (since y ( 0 ) = 1 ) . \int 3t\,dt = \frac{3t^2}{2} + C_2; \quad C_2=1 \text{ (since } y(0)=1 \text{)}. ∫ 3 t d t = 2 3 t 2 + C 2 ; C 2 = 1 (since y ( 0 ) = 1 ) . Third component: ∫ 4 e t d t = 4 e t + C 3 ; C 3 = 2 − 4 = − 2 (since z ( 0 ) = 2 ) . \int 4e^t\,dt = 4e^t + C_3; \quad C_3=2-4= -2 \text{ (since } z(0)=2 \text{)}. ∫ 4 e t d t = 4 e t + C 3 ; C 3 = 2 − 4 = − 2 (since z ( 0 ) = 2 ) . Therefore, r ⃗ ( t ) = ⟨ 2 t , 3 t 2 2 + 1 , 4 e t − 2 ⟩ . \vec{r}(t) = \langle 2t,\, \frac{3t^2}{2}+1,\, 4e^t-2 \rangle. r ( t ) = ⟨ 2 t , 2 3 t 2 + 1 , 4 e t − 2 ⟩ . 9.6 Solving Motion Problems Using Parametric and Vector–Valued Functions# Motion in the plane or space can be modeled using parametric or vector–valued functions. The derivative of the position function gives the velocity vector, and its magnitude yields speed. These representations are essential for solving problems in mechanics—for example, determining the path, speed, or acceleration of a moving object.
Self–Check 6# Problem:
r ⃗ ( t ) = ⟨ t , t 2 , ln ( t + 1 ) ⟩ . \vec{r}(t) = \langle t,\, t^2,\, \ln(t+1) \rangle. r ( t ) = ⟨ t , t 2 , ln ( t + 1 )⟩ . Determine the particle’s speed at time t = 1 t=1 t = 1
Solution:
First, find the velocity vector:
d d t ( t ) = 1 , \frac{d}{dt}(t)=1, d t d ( t ) = 1 , d d t ( t 2 ) = 2 t , \frac{d}{dt}(t^2)=2t, d t d ( t 2 ) = 2 t , d d t ( ln ( t + 1 ) ) = 1 t + 1 . \frac{d}{dt}(\ln(t+1))=\frac{1}{t+1}. d t d ( ln ( t + 1 )) = t + 1 1 . So,
v ⃗ ( t ) = ⟨ 1 , 2 t , 1 t + 1 ⟩ . \vec{v}(t) = \langle 1,\, 2t,\, \frac{1}{t+1} \rangle. v ( t ) = ⟨ 1 , 2 t , t + 1 1 ⟩ . At t = 1 t=1 t = 1
v ⃗ ( 1 ) = ⟨ 1 , 2 , 1 2 ⟩ . \vec{v}(1) = \langle 1,\, 2,\, \frac{1}{2} \rangle. v ( 1 ) = ⟨ 1 , 2 , 2 1 ⟩ . The speed is the magnitude:
∣ v ⃗ ( 1 ) ∣ = 1 2 + 2 2 + ( 1 2 ) 2 = 1 + 4 + 1 4 = 20 + 1 4 = 21 4 = 21 2 . |\vec{v}(1)| = \sqrt{1^2+2^2+\left(\frac{1}{2}\right)^2} = \sqrt{1+4+\frac{1}{4}} = \sqrt{\frac{20+1}{4}} = \sqrt{\frac{21}{4}} = \frac{\sqrt{21}}{2}. ∣ v ( 1 ) ∣ = 1 2 + 2 2 + ( 2 1 ) 2 = 1 + 4 + 4 1 = 4 20 + 1 = 4 21 = 2 21 . In polar coordinates, a point is defined by a radius r r r θ \theta θ
x = r cos θ , y = r sin θ . x = r\cos\theta, \quad y = r\sin\theta. x = r cos θ , y = r sin θ . When a curve is defined by a polar equation r = r ( θ ) r=r(\theta) r = r ( θ ) x x x
d y d x = d r d θ sin θ + r cos θ d r d θ cos θ − r sin θ . \frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}. d x d y = d θ d r cos θ − r sin θ d θ d r sin θ + r cos θ . This formula is essential for analyzing the slope of polar curves and for converting between coordinate systems.
Self–Check 7# Problem:
r ( θ ) = 2 + 2 sin θ , r(\theta)= 2+2\sin\theta, r ( θ ) = 2 + 2 sin θ , find d y d x \frac{dy}{dx} d x d y θ \theta θ
Solution:
Compute d r d θ = 2 cos θ . \frac{dr}{d\theta} = 2\cos\theta. d θ d r = 2 cos θ . Substitute into the formula: d y d x = ( 2 cos θ ) sin θ + ( 2 + 2 sin θ ) cos θ ( 2 cos θ ) cos θ − ( 2 + 2 sin θ ) sin θ . \frac{dy}{dx} = \frac{(2\cos\theta)\sin\theta + (2+2\sin\theta)\cos\theta}{(2\cos\theta)\cos\theta - (2+2\sin\theta)\sin\theta}. d x d y = ( 2 cos θ ) cos θ − ( 2 + 2 sin θ ) sin θ ( 2 cos θ ) sin θ + ( 2 + 2 sin θ ) cos θ . Simplify the numerator: = 2 cos θ sin θ + 2 cos θ + 2 sin θ cos θ 2 cos 2 θ − 2 sin θ − 2 sin 2 θ . = \frac{2\cos\theta\sin\theta + 2\cos\theta + 2\sin\theta\cos\theta}{2\cos^2\theta - 2\sin\theta - 2\sin^2\theta}. = 2 cos 2 θ − 2 sin θ − 2 sin 2 θ 2 cos θ sin θ + 2 cos θ + 2 sin θ cos θ . = 4 cos θ sin θ + 2 cos θ 2 cos 2 θ − 2 sin θ − 2 sin 2 θ . = \frac{4\cos\theta\sin\theta + 2\cos\theta}{2\cos^2\theta - 2\sin\theta - 2\sin^2\theta}. = 2 cos 2 θ − 2 sin θ − 2 sin 2 θ 4 cos θ sin θ + 2 cos θ . 9.8 Finding the Area of a Polar Region or the Area Bounded by a Single Polar Curve# The area of a region defined by a polar curve r = r ( θ ) r=r(\theta) r = r ( θ ) θ ∈ [ α , β ] \theta\in[\alpha,\beta] θ ∈ [ α , β ]
A = 1 2 ∫ α β [ r ( θ ) ] 2 d θ . A = \frac{1}{2}\int_{\alpha}^{\beta} [r(\theta)]^2\,d\theta. A = 2 1 ∫ α β [ r ( θ ) ] 2 d θ . This formula arises by dividing the region into narrow sectors and summing their areas.
Self–Check 8# Problem:
r ( θ ) = 3 cos θ , r(\theta)= 3\cos\theta, r ( θ ) = 3 cos θ , for θ \theta θ − π 2 -\frac{\pi}{2} − 2 π π 2 \frac{\pi}{2} 2 π
Solution:
Set up the area integral: A = 1 2 ∫ − π / 2 π / 2 [ 3 cos θ ] 2 d θ = 1 2 ∫ − π / 2 π / 2 9 cos 2 θ d θ . A = \frac{1}{2}\int_{-\pi/2}^{\pi/2} [3\cos\theta]^2\,d\theta = \frac{1}{2}\int_{-\pi/2}^{\pi/2} 9\cos^2\theta\,d\theta. A = 2 1 ∫ − π /2 π /2 [ 3 cos θ ] 2 d θ = 2 1 ∫ − π /2 π /2 9 cos 2 θ d θ . Factor constants: A = 9 2 ∫ − π / 2 π / 2 cos 2 θ d θ . A = \frac{9}{2}\int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta. A = 2 9 ∫ − π /2 π /2 cos 2 θ d θ . Use the identity cos 2 θ = 1 + cos 2 θ 2 \cos^2\theta = \frac{1+\cos2\theta}{2} cos 2 θ = 2 1 + c o s 2 θ A = 9 2 ⋅ 1 2 ∫ − π / 2 π / 2 ( 1 + cos 2 θ ) d θ = 9 4 [ θ + sin 2 θ 2 ] − π / 2 π / 2 . A = \frac{9}{2}\cdot\frac{1}{2}\int_{-\pi/2}^{\pi/2} (1+\cos2\theta)\,d\theta = \frac{9}{4}\left[\theta + \frac{\sin2\theta}{2}\right]_{-\pi/2}^{\pi/2}. A = 2 9 ⋅ 2 1 ∫ − π /2 π /2 ( 1 + cos 2 θ ) d θ = 4 9 [ θ + 2 sin 2 θ ] − π /2 π /2 . Evaluate:At θ = π / 2 \theta=\pi/2 θ = π /2 π 2 + sin π 2 = π 2 . \frac{\pi}{2} + \frac{\sin\pi}{2} = \frac{\pi}{2}. 2 π + 2 s i n π = 2 π . At θ = − π / 2 \theta=-\pi/2 θ = − π /2 − π 2 + sin ( − π ) 2 = − π 2 . -\frac{\pi}{2} + \frac{\sin(-\pi)}{2} = -\frac{\pi}{2}. − 2 π + 2 s i n ( − π ) = − 2 π . A = 9 4 ( π 2 − ( − π 2 ) ) = 9 4 ⋅ π = 9 π 4 . A = \frac{9}{4}\left(\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right) = \frac{9}{4}\cdot \pi = \frac{9\pi}{4}. A = 4 9 ( 2 π − ( − 2 π ) ) = 4 9 ⋅ π = 4 9 π . 9.9 Finding the Area of the Region Bounded by Two Polar Curves# When two polar curves, r = r 1 ( θ ) r=r_1(\theta) r = r 1 ( θ ) r = r 2 ( θ ) r=r_2(\theta) r = r 2 ( θ ) r 1 ( θ ) ≥ r 2 ( θ ) r_1(\theta) \ge r_2(\theta) r 1 ( θ ) ≥ r 2 ( θ )
A = 1 2 ∫ α β ( [ r 1 ( θ ) ] 2 − [ r 2 ( θ ) ] 2 ) d θ . A = \frac{1}{2}\int_{\alpha}^{\beta} \left( [r_1(\theta)]^2 - [r_2(\theta)]^2 \right)\,d\theta. A = 2 1 ∫ α β ( [ r 1 ( θ ) ] 2 − [ r 2 ( θ ) ] 2 ) d θ . This method subtracts the inner area from the outer area to obtain the net region.
Self–Check 9# Problem:
r 1 ( θ ) = 4 , r 2 ( θ ) = 2 , r_1(\theta)= 4, \quad r_2(\theta)= 2, r 1 ( θ ) = 4 , r 2 ( θ ) = 2 , for θ \theta θ 0 0 0 2 π 2\pi 2 π
Solution:
Since the curves are circles (with constant radii), the area between them is: A = 1 2 ∫ 0 2 π ( 4 2 − 2 2 ) d θ = 1 2 ∫ 0 2 π ( 16 − 4 ) d θ = 1 2 ∫ 0 2 π 12 d θ . A = \frac{1}{2}\int_0^{2\pi} \left(4^2 - 2^2\right)\,d\theta = \frac{1}{2}\int_0^{2\pi} \left(16-4\right)\,d\theta = \frac{1}{2}\int_0^{2\pi} 12\,d\theta. A = 2 1 ∫ 0 2 π ( 4 2 − 2 2 ) d θ = 2 1 ∫ 0 2 π ( 16 − 4 ) d θ = 2 1 ∫ 0 2 π 12 d θ . Integrate: A = 1 2 ⋅ 12 ⋅ ( 2 π ) = 12 π . A = \frac{1}{2}\cdot 12\cdot (2\pi) = 12\pi. A = 2 1 ⋅ 12 ⋅ ( 2 π ) = 12 π . 
