Consider the curves in the first quadrant that have equations y = A exp(7x), whe
ID: 2875383 • Letter: C
Question
Consider the curves in the first quadrant that have equations y = A exp(7x), where A is a positive constant. Different values of A give different curves. The curves form a family, F. Let P = (5, 4). Let C be the member of the family F that goes through P. Let y = f(x) be the equation of C. Find f(x). f(x) = Find the slope at P of the tangent to C. slope = A curve D is perpendicular to C at P. What is the slope of the tangent to D at the point P? slope = Give a formula g(y) 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) = 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) =Explanation / Answer
a> the curve C passes through the point P(5,4)
=> the equation of the curve C will satisty the point P
=> y=f(x) =Ae^(7x)
at the point P
f(5) = 4 = Ae^(7*5) = Ae^(35) , A = 4/e^(35) or A = 4e^(-35)
=> f(x) = 4e^(-35)*e^(7x)
or f(x) = 4e^(7x - 35)
b> the slope of C at the point P is the derivative of f(x) at P
=> f'(x) = 4e^(7x-35)*7 = 28e^(7x-35)
f'(5) = 28e^(35-35) = 28
=> the slope at the point P(5,4) is = 28