190 STEP 4. Review the Knowledge You Need to Score High
9.4 Parametric, Polar, and Vector Derivatives
Main Concepts:Derivatives of Parametric Equations; Position, Speed, and
Acceleration; Derivatives of Polar Equations; Velocity
and Acceleration of Vector FunctionsDerivatives of Parametric Equations
If a function is defined parametrically, you can differentiate bothx(t) andy(t) with respect
tot, and thendy
dx=
dy
dt÷
dx
dt=
dy
dt·
dt
dx.
Example 1
A curve is defined byx(t)=t^2 − 3 tandy(t)=5 cost. Find
dy
dx.
Step 1: Differentiatex(t) andy(t) with respect tot.
dx
dt
= 2 t−3 and
dy
dt
=−5 sint.Step 2:
dy
dx=
dy
dt·
dt
dx=
−5 sint
2 t− 3Example 2
A function is defined byx(t)= 5 t−2 andy(t)= 9 −t^2 when− 5 ≤t≤5. Find the equation
of any horizontal tangent lines to the curve.Step 1: Differentiatex(t) andy(t) with respect tot.
dx
dt=5 and
dy
dt=− 2 t.Step 2:
dy
dx=
dy
dt·
dt
dx=
− 2 t
5Step 3: In order for the tangent line to be horizontal,
dy
dx
must be equal to zero, therefore
t=0,x=−2, andy=9.
Step 4: The equation of the horizontal tangent line at (−2, 9) isy=9.Example 3
A curve is defined byx(t)=t^2 − 5 t+2 andy(t)=1
(t+2)^2
for 0≤t≤3. Find the equation
of the tangent line to the curve whent=1.Step 1:
dx
dt
= 2 t−5 and
dy
dt=
− 2
(t+2)^3.
Step 2:
dy
dx=
− 2
(t+2)^3·
1
(2t−5)Step 3: Att=1,m=− 2
(3)^3
·
1
(−3)
=
2
81
,x= 1 − 5 + 2 =−2, andy=1
(3)^2
=
1
9
.
Step 4: The equation of the tangent line isy−1
9
=
2
81
(x+2) ory=2
81
x+