High School
SAT
SAT Elite 1500
SAT Tutoring
ACT
ACT Elite 33
ACT Tutoring
University
MCAT
MCAT Elite 515
Med-School Admissions
Pre-Med Tutoring
Pre-Med Plus
LSAT
LSAT Elite 170
LSAT Self-Paced
LSAT Tutoring
DAT
DAT Elite
DAT Tutoring
Log in
Get Started for Free
Q: Find the arc length of y=2x^3/2 on [0, 5/3].
Related Topics
Wize University Calculus 1 Textbook > Applications of Integration
Arc Length
3 Activities
Q
:
\bf{Q:}
Q
:
Find the arc length of
y
=
2
x
3
2
y=2x^{\frac{3}{2}}
y
=
2
x
2
3
on
[
0
,
5
3
]
\left[0,\ \dfrac{5}{3}\right]
[
0
,
3
5
]
.
Answer
I don't know
Check Submission
More Arc Length Questions:
Arc Length
Which of the following integrals represent the arc length of the curve
r
=
sin
2
(
θ
4
)
r=\sin^2\left(\frac{\theta}{4}\right)
r
=
sin
2
(
4
θ
)
?
Find the arc length of
y
=
2
x
3
2
y=2x^{\frac{3}{2}}
y
=
2
x
2
3
on
[
0
,
5
3
]
\left[0,\ \dfrac{5}{3}\right]
[
0
,
3
5
]
.
Find the arc length of the curve
y
=
3
+
1
2
cosh
2
x
y = 3 +\dfrac{1}{2}\cosh\ 2x
y
=
3
+
2
1
cosh
2
x
from
x
=
0
x=0
x
=
0
to
x
=
1
x=1
x
=
1
.
Function from Arc Length
Find a curve through the point
(
0
,
1
)
(0, 1)
(
0
,
1
)
whose length integral is
L
=
∫
1
2
1
+
1
y
4
d
y
\displaystyle L=\int_1^2\sqrt{1+\frac{1}{y^4}}\ dy
L
=
∫
1
2
1
+
y
4
1
d
y
Arc Length with Partial Fractions
Find the arc length of
y
=
e
x
y=e^x
y
=
e
x
on
[
0
,
1
]
[0,1]
[
0
,
1
]
.
Practice: Arc Length with Perfect Square
Q:
\textbf{Q:}
Q:
Consider the curve
y
=
f
(
x
)
=
x
3
3
+
1
4
x
y=f(x)=\dfrac{x^3}{3}+\dfrac{1}{4x}
y
=
f
(
x
)
=
3
x
3
+
4
x
1
. Find the arc length of this curve over the interval
[
1
,
3
]
[1, 3]
[
1
,
3
]
. Find a function giving the arc length of the curve over the interval
[
1
,
x
]
.
[1, x].
[
1
,
x
]
.
Find the arc length of
y
=
2
x
3
2
y=2x^{\frac{3}{2}}
y
=
2
x
2
3
on
[
0
,
5
3
]
\left[0,\ \dfrac{5}{3}\right]
[
0
,
3
5
]
.
(a)
5
2
5\sqrt2
5
2
(b)
2
3
\frac23
3
2
(c)
14
3
\frac{14}{3}
3
14
(d)
7
3
7\sqrt3
7
3
(e)
2
27
\frac{2}{27}
27
2
Q
:
\bf{Q:}
Q
:
Find the arc length of
y
=
2
x
3
2
y=2x^{\frac{3}{2}}
y
=
2
x
2
3
on
[
0
,
5
3
]
\left[0,\ \dfrac{5}{3}\right]
[
0
,
3
5
]
.
Arc Length
Which of the following integrals represent the arc length of the curve
r
=
sin
2
(
θ
4
)
r=\sin^2\left(\frac{\theta}{4}\right)
r
=
sin
2
(
4
θ
)
?
Practice Question: Arc Length
Find the arc length function for the curve
y
=
2
x
3
/
2
y=2x^{3/2}
y
=
2
x
3/2
with starting point
P
0
=
(
0
,
0
)
P_0=\left(0,0\right)
P
0
=
(
0
,
0
)
.
Q
:
\bf{Q:}
Q
:
Find the arc length of
y
=
2
x
3
2
y=2x^{\frac{3}{2}}
y
=
2
x
2
3
on
[
0
,
5
3
]
\left[0,\ \dfrac{5}{3}\right]
[
0
,
3
5
]
.
Find the arc length of the curve
y
=
3
+
1
2
cosh
2
x
y = 3 +\dfrac{1}{2}\cosh\ 2x
y
=
3
+
2
1
cosh
2
x
from
x
=
0
x=0
x
=
0
to
x
=
1
x=1
x
=
1
.
Compute the arc length of the curve
f
(
x
)
=
∫
1
x
t
5
−
1
d
t
f\left(x\right)=\int_1^x\sqrt{t^5-1}dt\
f
(
x
)
=
∫
1
x
t
5
−
1
d
t
for
1
≤
x
≤
9
\ 1\le x\le9
1
≤
x
≤
9
.
Hint: use the Fundamental Theorem of Calculus:
d
d
x
∫
c
x
f
(
t
)
d
t
=
f
(
x
)
\frac{d}{dx}\int_c^x f(t)dt=f(x)
d
x
d
∫
c
x
f
(
t
)
d
t
=
f
(
x
)
Arc Length with Partial Fractions
Find the arc length of
y
=
e
x
y=e^x
y
=
e
x
on
[
0
,
1
]
[0,1]
[
0
,
1
]
.
Function from Arc Length
Find a curve through the point
(
0
,
1
)
(0, 1)
(
0
,
1
)
whose length integral is
L
=
∫
1
2
1
+
1
y
4
d
y
\displaystyle L=\int_1^2\sqrt{1+\frac{1}{y^4}}\ dy
L
=
∫
1
2
1
+
y
4
1
d
y
Practice: Arc Length with Perfect Square
Q:
\textbf{Q:}
Q:
Consider the curve
y
=
f
(
x
)
=
x
3
3
+
1
4
x
y=f(x)=\dfrac{x^3}{3}+\dfrac{1}{4x}
y
=
f
(
x
)
=
3
x
3
+
4
x
1
. Find the arc length of this curve over the interval
[
1
,
3
]
[1, 3]
[
1
,
3
]
. Find a function giving the arc length of the curve over the interval
[
1
,
x
]
.
[1, x].
[
1
,
x
]
.