Respuesta :
Answer:
The distance travelled by the kite is 122.8 ft ( approx )
Step-by-step explanation:
Here, the given function,
[tex]y=150-\frac{1}{40}(x-50)^2[/tex]
Differentiating with respect to x,
[tex]y'=-\frac{1}{20}(x-50)[/tex]
∵ arc length of a curve is,
[tex]L=\int_{a}^{b} \sqrt{1+y'^2}dx[/tex]
Where, y shows the height of the curve for a ≤ x ≤ b,
Thus, the arc length of the given curve is,
[tex]L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx[/tex]
Put [tex]-\frac{1}{20}(x-50)=tan\theta[/tex]
[tex]\implies -dx=-20 sec^2\theta d\theta[/tex]
[tex]\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta[/tex]
[tex]=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta[/tex]
[tex]=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta[/tex]
By integration by parts,
[tex]=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}[/tex]
If x = 80, [tex]tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}[/tex]
[tex]sec \theta = \frac{\sqrt{13}}{2}[/tex]
[tex]\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}[/tex]
[tex]sec \theta = \frac{\sqrt{29}}{2}[/tex]
Thus, the length of the curve is,
[tex]=-10(\frac{\sqrt{13}}{2}(-\frac{3}{2}) +ln|\frac{13}{2}-\frac{3}{2}|) + 10(\frac{5\sqrt{29}}{4} + ln |\frac{29}{2} + \frac{5}{2} |)[/tex]
[tex]\approx 122.8\text{ feet}[/tex]
The distance between the point (0, 87.5) and (80, 127.5). Then the distance traveled by the kite is 89.44 ft.
What is the distance?
It is the measure of length between the two points.
The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by
[tex]\rm y = 150 -\dfrac{1}{ 40} (x - 50)^2[/tex]
For x = 0, the value of y will be
[tex]\rm y = 150 - \dfrac{1}{40} (0 - 50)^2\\\\y = 150 - \dfrac{1 }{40}*2500\\\\y = 150 - 62.5\\\\y = 87.5[/tex]
For x = 80, the value of y will be
[tex]\rm y = 150 - \dfrac{1}{40} (80 - 50)^2\\\\y = 150 - \dfrac{1 }{40}*900\\\\y = 150 - 22.5\\\\y = 12.75[/tex]
Then we have the distance formula,
[tex]\rm D = \sqrt{(80-0)^2+(127.5-87.5)^2}\\\\D = \sqrt{6400 + 1600}\\\\D =\sqrt{8000}\\\\D = 89.44[/tex]
Thus, the distance between the kite and the starting point is 89.44 ft.
More about the Distance link is given below.
https://brainly.com/question/989117
