Second Derivatives of Parametric Equations
AP Calculus BC · Flashcards · Application
AP CALCULUS BC · FLASHCARDS
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QUESTION
Calculate
d
y
d
x
\frac{dy}{dx}
d
x
d
y
when
y
(
t
)
=
t
2
+
1
y(t) = t^2 + 1
y
(
t
)
=
t
2
+
1
and
x
(
t
)
=
3
t
x(t) = 3t
x
(
t
)
=
3
t
.
QUESTION
Find
d
x
d
t
\frac{dx}{dt}
d
t
d
x
for
x
(
t
)
=
2
t
2
+
3
t
x(t) = 2t^2 + 3t
x
(
t
)
=
2
t
2
+
3
t
.
QUESTION
What is
d
y
d
x
\frac{dy}{dx}
d
x
d
y
for
y
(
t
)
=
t
2
y(t) = t^2
y
(
t
)
=
t
2
and
x
(
t
)
=
t
2
x(t) = t^2
x
(
t
)
=
t
2
?
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ANSWER
1
1
1
. Since both functions are identical,
d
y
d
x
=
2
t
2
t
=
1
\frac{dy}{dx} = \frac{2t}{2t} = 1
d
x
d
y
=
2
t
2
t
=
1
.
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