Finding "curve"

Calculus Level 4

Let the equation of the curve $y=f(x)$ is such that $\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\sqrt { 1-{ \left( \dfrac { dy }{ dx } \right) }^{ 2 } }$ and the tangent to the curve at the origin is inclined at $\dfrac{ \pi } { 4}$ with the postive direction of the $x$ -axis. Find the value of $y$ at $x= \dfrac { \pi }{2}$

NOTE:- $f^{ ' }\left( x \right) \neq constant$

The answer is 1.000.

1 solution

Tom Engelsman
Jun 28, 2016

Let u = dy/dx so that one obtains the first-order differential equation:

du/dx = sqrt(1 - u^2);

which separates into:

du/sqrt(1 - u^2) = dx;

and integrates into:

arcsin(u) = x + A => u = sin(x + A) => dy/dx = sin(x + A)

where A is a real constant of integration. If the tangent to y(x) at the origin is inclined 45 degrees with respect to the positive x-axis, then we have the boundary condition y' (0) = tan(pi/4) = 1. Solving for A gives:

1 = sin(0 + A) => A = pi/2.

hence, we arrive at dy/dx = sin(x + pi/2). A final integration gives y(x) = -cos(x + pi/2) + B, which we are told the origin, or y(0) = 0, is included on the graph and yields B = 0.

The required function is just y(x) = -cos(x + pi/2), and y(pi/2) = 1.

Finding "curve"

The answer is 1.000.

1 solution

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