Answer:
(5) - Option C, [tex]s=3\sqrt{17}[/tex]
(6) - Option D, [tex]-\frac{1}{4} y^{-4}=\frac{1}{3} x^3 -\frac{1}{4}[/tex]
Step-by-step explanation:
Given the following questions.
(5) - Find the arc length of y=4x-3 from A(1,1) to B(4,13)
(6) - Solve the first-order differential equation [tex]y'=x^2y^5[/tex] with the initial condition, [tex]y(0)=1[/tex].
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Question #5:
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Formula for Arc Length:}}\\\\s=\int\limits^b_a {\sqrt{1+(f'(x))^2} } \, dx \end{array}\right}[/tex]
(1) - Take the derivative of the function y
[tex]y=4x-3\\\\\Longrightarrow \boxed{y'=4}[/tex]
(2) - Square y'
[tex]y'=4\\\\\Longrightarrow y'=4^2\\\\\Longrightarrow \boxed{y'=16}[/tex]
(3) - Plug into the formula for arc length
[tex]s=\int\limits^b_a {\sqrt{1+(f'(x))^2} } \, dx \\\\\text{Limits:} \ 1\leq x\leq 4\\\\\Longrightarrow\int\limits^4_1 {\sqrt{1+16} } \, dx \\\\\Longrightarrow\boxed{ \int\limits^4_1 {\sqrt{17} } \, dx} \\[/tex]
(4) - Solve the integral
[tex]\int\limits^4_1 {\sqrt{17} } \, dx \\\\\Longrightarrow \Big [x\sqrt{17} \Big] \right]_{1}^{4}\\\\\Longrightarrow 4\sqrt{17} -\sqrt{17\\}\\\\ \therefore \boxed{s=3\sqrt{17} }[/tex]
Thus, the arc length is found.
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Quick note: The question solved for the arc length using a dy integral not a dx integral (which was really unnecessary), so let me clarify that issue.
(1) - Taking the function y, and solving it for x
[tex]y=4x-3\\\\\Longrightarrow y+3=4x\\\\\Longrightarrow \boxed{x=\frac{1}{4}y+\frac{3}{4}}[/tex]
(2) - Repeating steps (1)-(4) from above
[tex]x=\frac{1}{4}y+\frac{3}{4}\\\\\Longrightarrow \boxed{x'=\frac{1}{4}} \\\\\Longrightarrow (x')^2=(\frac{1}{4})^2\\\\\Longrightarrow \boxed{(x')^2=\frac{1}{16}}\\\\s=\int\limits^b_a {\sqrt{1+(f'(x))^2} } \, dy\\\text{Limits:} \ 1\leq y\leq 13\\\\ \Longrightarrow\int\limits^{13}_1 {\sqrt{1+\frac{1}{16} } \, dy\\\\ \Longrightarrow\int\limits^{13}_1 {\sqrt{\frac{17}{16} } \, dy\\\\ \Longrightarrow\int\limits^{13}_1 {\frac{\sqrt{17} }{4} } \, dy\\\\[/tex]
[tex]\Longrightarrow\Big[\frac{\sqrt{17} }{4} y \Big]^{13}_{1}\\\\\Longrightarrow \frac{13\sqrt{17} }{4} -\frac{\sqrt{17} }{4} \\\\\ \Longrightarrow \boxed{\boxed{s=3\sqrt{17} }}[/tex]
Thus, the correct setup according to your question is option C.
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Question #6:
The given differential equation is separable.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Seperable Differential Equation:}}\\\frac{dy}{dx} =f(x)g(y)\\\\\rightarrow\int\frac{dy}{g(y)}=\int f(x)dx \end{array}\right }[/tex]
(1) - Solve the separable DE
[tex]\frac{dy}{dx} =x^2y^5\\\\\Longrightarrow \frac{1}{y^5} dy=x^2dx\\\\\Longrightarrow \int y^{-5}dy= \int x^2 dx\\\\\Longrightarrow \boxed{ -\frac{1}{4} y^{-4}=\frac{1}{3} x^3 +C}[/tex]
(2) - Use the given initial condition to find the arbitrary constant "C"
[tex]\text{Recall} \rightarrow y(0)=1\\\\-\frac{1}{4} y^{-4}=\frac{1}{3} x^3 +C\\\\\Longrightarrow -\frac{1}{4} (1)^{-4}=\frac{1}{3} (0)^3 +C\\\\\Longrightarrow -\frac{1}{4} (1)=0 +C\\\\\therefore \boxed{C=-\frac{1}{4} }[/tex]
(3) - Form the final solution
[tex]\boxed{\boxed{-\frac{1}{4} y^{-4}=\frac{1}{3} x^3 -\frac{1}{4} }}[/tex]
Thus, option D is correct.
