Let
x
(
t
)
=
t
3
−
4
t
2
+
3
t
\displaystyle {x}{\left({t}\right)}={t}^{{3}}-{4}{t}^{{2}}+{3}{t}
x
(
t
)
=
t
3
−
4
t
2
+
3
t
and
y
(
t
)
=
t
2
+
1
\displaystyle {y}{\left({t}\right)}={t}^{{2}}+{1}
y
(
t
)
=
t
2
+
1
At
t
=
1
\displaystyle {t}={1}
t
=
1
,
x
(
1
)
=
\displaystyle {x}{\left({1}\right)}=
x
(
1
)
=
y
(
1
)
=
\displaystyle {y}{\left({1}\right)}=
y
(
1
)
=
d
x
d
t
∣
t
=
1
=
\displaystyle \frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}{\mid}_{{{t}={1}}}=
d
t
d
x
∣
t
=
1
=
d
y
d
t
∣
t
=
1
=
\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}{\mid}_{{{t}={1}}}=
d
t
d
y
∣
t
=
1
=
d
y
d
x
∣
t
=
1
\displaystyle \frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}{\mid}_{{{t}={1}}}
d
x
d
y
∣
t
=
1
= tangent slope =
speed(1) =
\displaystyle \text{speed(1) =}
speed(1) =
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