Question
This problem is solved without a calculator. The slope of a function f at any given point (x, y) is \(\frac{2y}{3x^{2}}\) . The point (3, 4) is on the graph of f.
(A) Write an equation of the tangent line to the graph of f at x = 3.
(B) Use the tangent line in part (a) to approximate f (5). Is this approximation an overestimate or underestimate? Justify your answer using the second derivative test.
(C) Solve the separable differential equation \( \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y}{3x^{2}}\) with initial condition f(3) = 4.
(D) Use the solution in part (c) to find f(5).
Answer/Explanation
(A) The slope at x=3 (i.e.,the point (3,4)) is
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y}{3x^{2}}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(4)}{3(3)^{2}}=\frac{8}{27}\)
Now write the equation of the tangent line at (3, 4) in point-slope form.
\(y-y_{0}=m(x-x_{0})\Rightarrow y-4=\frac{8}{27}(x-3)\Rightarrow y=\frac{8}{27}(x-3)+4\Rightarrow y=\frac{8}{27}x+\frac{28}{9}\)
(B)\(f(5)=\frac{8}{27}(5-3)+4=\frac{124}{27}\) This is an overestimate because the graph is concavedown over the interval (3,5) with a negative second derivative of \(\left ( (3x^{2})y’-6x(2y) \right )/(3x^{2})^{2}\)
(C) Start by separating the variables and integrating both sides
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2y}{3x^{2}}\Rightarrow \frac{1}{2y}dy=\frac{1}{3x^{2}}dx\Rightarrow \int \frac{1}{2y}dy=\int \frac{1}{3x^{2}}dx
\frac{1}{2}\int \frac{1}{y}dy=\frac{1}{3}\int x^{-2}dx\Rightarrow \frac{1}{2}\ln |y|=\left ( \frac{1}{3} \right )\frac{x^{-1}}{-1}+c_{1}\Rightarrow \frac{1}{2}\ln \left | y \right |=-\frac{1}{3x}+c_{1}
\ln \left | y \right |=-\frac{2}{3x}+c_{2}\Rightarrow e^{\ln \left | y \right |}=e^{-\frac{2}{3x}+c_{2}}=e^{-\frac{2}{3x}.e^{c_{2}}}\)
\(y=c_{3}e^{-\frac{2}{3x}}\)
Now use the initial condition f(3) = 4 to solve for \(c_{3}\).
\(4=c_{3}e^{-\frac{2}{3(3)}}\Rightarrow c_{3}=4.9954\)
\(y=4.9954e^{-\frac{2}{3x}}\)
(D) Evaluate the equation in C at f(5).
\(f(5)=4.9954e^{-\frac{2}{3(5)}}\approx 4.37184\)