Answer:
The are a is 6
The length of h = 2.4
Step-by-step explanation:
This is a 3 4 5 special triangle so the length of the hypotenuse is 5
The area for a triangle is calculated by multiplying height to the base and that divided by two
Since it's a right triangle we can take any side length as the height and the result wouldn't change
4×3÷2 = 6 is the area for this triangle now we can use thus information to find the length of h :
h × 5 ÷ 2 = 6
5h = 12
h = 2.4
(5b, -1z)
You have to multiply the B and the C by 5 to get the original equation or you could divide the original equation by 5
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=6The arclength of a parametric curve can be found using the formula:
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=
6
√
3
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
