To solve these problems based on the parametric curve:
x ( t ) = sin ( t 4 ) + cos ( t ) , y ( t ) = cos ( t 4 ) − sin ( t ) , 0 ≤ t ≤ π 2 , x(t) = \sin\left(\frac{t}{4}\right) + \cos(t), \quad y(t) = \cos\left(\frac{t}{4}\right) - \sin(t), \quad 0 \leq t \leq \frac{\pi}{2}, x ( t ) = sin ( 4 t ) + cos ( t ) , y ( t ) = cos ( 4 t ) − sin ( t ) , 0 ≤ t ≤ 2 π ,
we need to set up the appropriate integrals for:
The length of the curve.
The surface area obtained when the curve is revolved around the y y y
(a) Length of the curve The formula for the length of a parametric curve x ( t ) , y ( t ) x(t), y(t) x ( t ) , y ( t ) [ a , b ] [a, b] [ a , b ]
s = ∫ a b ( d x d t ) 2 + ( d y d t ) 2 d t . s = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. s = ∫ a b ( d t d x ) 2 + ( d t d y ) 2 d t .
Steps:
Find d x d t \frac{dx}{dt} d t d x
Differentiate x ( t ) = sin ( t 4 ) + cos ( t ) x(t) = \sin\left(\frac{t}{4}\right) + \cos(t) x ( t ) = sin ( 4 t ) + cos ( t ) d x d t = 1 4 cos ( t 4 ) − sin ( t ) . \frac{dx}{dt} = \frac{1}{4} \cos\left(\frac{t}{4}\right) - \sin(t). d t d x = 4 1 cos ( 4 t ) − sin ( t ) .
Find d y d t \frac{dy}{dt} d t d y
Differentiate y ( t ) = cos ( t 4 ) − sin ( t ) y(t) = \cos\left(\frac{t}{4}\right) - \sin(t) y ( t ) = cos ( 4 t ) − sin ( t ) d y d t = − 1 4 sin ( t 4 ) − cos ( t ) . \frac{dy}{dt} = -\frac{1}{4} \sin\left(\frac{t}{4}\right) - \cos(t). d t d y = − 4 1 sin ( 4 t ) − cos ( t ) .
Square and add the derivatives:
( d x d t ) 2 + ( d y d t ) 2 = ( 1 4 cos ( t 4 ) − sin ( t ) ) 2 + ( − 1 4 sin ( t 4 ) − cos ( t ) ) 2 . \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \left( \frac{1}{4} \cos\left(\frac{t}{4}\right) - \sin(t) \right)^2 + \left( -\frac{1}{4} \sin\left(\frac{t}{4}\right) - \cos(t) \right)^2. ( d t d x ) 2 + ( d t d y ) 2 = ( 4 1 cos ( 4 t ) − sin ( t ) ) 2 + ( − 4 1 sin ( 4 t ) − cos ( t ) ) 2 .
Integrate:
Substitute this expression into the length formula:
s = ∫ 0 π 2 ( 1 4 cos ( t 4 ) − sin ( t ) ) 2 + ( − 1 4 sin ( t 4 ) − cos ( t ) ) 2 d t . s = \int_0^{\frac{\pi}{2}} \sqrt{\left( \frac{1}{4} \cos\left(\frac{t}{4}\right) - \sin(t) \right)^2 + \left( -\frac{1}{4} \sin\left(\frac{t}{4}\right) - \cos(t) \right)^2} \, dt. s = ∫ 0 2 π ( 4 1 cos ( 4 t ) − sin ( t ) ) 2 + ( − 4 1 sin ( 4 t ) − cos ( t ) ) 2 d t .
(b) Surface area of revolution around the y y y The formula for the surface area of a parametric curve x ( t ) , y ( t ) x(t), y(t) x ( t ) , y ( t ) y y y
S = ∫ a b 2 π x ( t ) ( d x d t ) 2 + ( d y d t ) 2 d t . S = \int_a^b 2\pi x(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. S = ∫ a b 2 π x ( t ) ( d t d x ) 2 + ( d t d y ) 2 d t .
Steps:
From part (a), we already have the expression for:
( d x d t ) 2 + ( d y d t ) 2 . \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}. ( d t d x ) 2 + ( d t d y ) 2 .
The function x ( t ) x(t) x ( t ) x ( t ) = sin ( t 4 ) + cos ( t ) x(t) = \sin\left(\frac{t}{4}\right) + \cos(t) x ( t ) = sin ( 4 t ) + cos ( t )
Integrate:
Substitute into the surface area formula:
S = ∫ 0 π 2 2 π ( sin ( t 4 ) + cos ( t ) ) ( 1 4 cos ( t 4 ) − sin ( t ) ) 2 + ( − 1 4 sin ( t 4 ) − cos ( t ) ) 2 d t . S = \int_0^{\frac{\pi}{2}} 2\pi \left( \sin\left(\frac{t}{4}\right) + \cos(t) \right) \sqrt{\left( \frac{1}{4} \cos\left(\frac{t}{4}\right) - \sin(t) \right)^2 + \left( -\frac{1}{4} \sin\left(\frac{t}{4}\right) - \cos(t) \right)^2} \, dt. S = ∫ 0 2 π 2 π ( sin ( 4 t ) + cos ( t ) ) ( 4 1 cos ( 4 t ) − sin ( t ) ) 2 + ( − 4 1 sin ( 4 t ) − cos ( t ) ) 2 d t .
Final Answers:
Curve length integral (a):
s = ∫ 0 π 2 ( 1 4 cos ( t 4 ) − sin ( t ) ) 2 + ( − 1 4 sin ( t 4 ) − cos ( t ) ) 2 d t . s = \int_0^{\frac{\pi}{2}} \sqrt{\left( \frac{1}{4} \cos\left(\frac{t}{4}\right) - \sin(t) \right)^2 + \left( -\frac{1}{4} \sin\left(\frac{t}{4}\right) - \cos(t) \right)^2} \, dt. s = ∫ 0 2 π ( 4 1 cos ( 4 t ) − sin ( t ) ) 2 + ( − 4 1 sin ( 4 t ) − cos ( t ) ) 2 d t .
Surface area around the y y y
S = ∫ 0 π 2 2 π ( sin ( t 4 ) + cos ( t ) ) ( 1 4 cos ( t 4 ) − sin ( t ) ) 2 + ( − 1 4 sin ( t 4 ) − cos ( t ) ) 2 d t . S = \int_0^{\frac{\pi}{2}} 2\pi \left( \sin\left(\frac{t}{4}\right) + \cos(t) \right) \sqrt{\left( \frac{1}{4} \cos\left(\frac{t}{4}\right) - \sin(t) \right)^2 + \left( -\frac{1}{4} \sin\left(\frac{t}{4}\right) - \cos(t) \right)^2} \, dt. S = ∫ 0 2 π 2 π ( sin ( 4 t ) + cos ( t ) ) ( 4 1 cos ( 4 t ) − sin ( t ) ) 2 + ( − 4 1 sin ( 4 t ) − cos ( t ) ) 2 d t .
Follow-Up Questions:
How do you compute derivatives of parametric equations?
What is the general formula for the length of a parametric curve?
How do you derive the surface area for a curve revolved around the x x x y y y
Can you calculate these integrals numerically?
What are the applications of parametric curves in physics or engineering?
Tip: Always differentiate parametric equations carefully and substitute into formulas step by step to avoid errors.