Question

This question is not from a test. This is an extra credit MATHS assignment. Please help me if you can. If you do, thank you! :)

Answer 1

we have the equation

`y=31sin5x+32`

Find out the first derivative

`\begin{gathered} y^(\prime)=(5)(31)c0s5x \\ y^(\prime)=155cos5x \end{gathered}`

equate to zero the derivative, to find out the critical points

`\begin{gathered} 155cos5x=0 \\ cos5x=0 \end{gathered}`

The value of cosine is zero when the angle is 90, 270 degrees (pi/2 and 3pi/2)

so

`\begin{gathered} 5x=(\pi)/(2) \\ \\ x=(\pi)/(10) \end{gathered}`
`\begin{gathered} 5x=(3\pi)/(2) \\ \\ x=(3\pi)/(10) \end{gathered}`

For x=pi/10

Find out the y-coordinate

`\begin{gathered} y=31s\imaginaryI n(5\pi)/(10)+32 \\ \\ y=31sin(\pi)/(2)+32 \\ \\ y=31+32=63\text{ ft} \end{gathered}`

Verify for x=3pi/10

`\begin{gathered} y=31s\imaginaryI n5((3\pi)/(10))+32 \\ \\ y=31s\imaginaryI n((3\pi)/(2))+32 \\ y=-31+32=1\text{ ft} \end{gathered}`

The maximum height is 63 ft