Consider the curves in the first quadrant that have equation

Consider the curves in the first quadrant that have equations y = A exp(4x), where A is a positive constant. Different values of A give different curves. The curves form a family, F. Let P = (7, 6). Let C be the member of the family F that goes through P. A. Let y = f(x) be the equation of C. Find f(x). f(x) = B. Find the slope at P of the tangent to C. slope = C. A curve D is perpendicular to C at P. What is the slope of the tangent to D at the point P? slope = D. Give a formula for the slope at (x, y) of the member of F that goes through (x, y). The formula should not involve A or x. g(y)= E. A curve which at each of its points is perpendicular to the member of the family F that goes through that point is called an orthogonal trajectory to F. Each orthogonal trajectory to F satisfies the differential equation dy/dx = -1/g(y), where g(y) is the answer to part D. Find a function h(y) such that x = h(y) is the equation of the orthogonal trajectory to F that passes through the point P. h(y) =

Solution

dy/dx=-1/g(y)

dy/dx=-1/(4y)

4ydy=-dx

INtegrating gives

-x=2y^2+C

x=-2y^2-C

It passes through P(7,6)

So,

7=-2*6^2-C

-C=7+2*36=79

x=-2y^2+79

h(y)=-2y^2+79