1. ## Find the equation

Consider the curves in the first quadrant that have equations
where is a positive constant.
Different values of give different curves. The curves form a family, .
Let Let be the member of the family that goes through P.
A. Let be the equation of . Find .

2. Well? You are told that the equation is of the form y= Aexp(4x) and that its graph goes through (2, 2). y= 2= A exp(4(2))= A exp(8). what is A?

3. ln(2)/8???

4. Originally Posted by amiv4
ln(2)/8???
$A = \frac{2}{\exp (8)}$

so,

$f(x) = \frac{2}{\exp (8)} \exp(4x)$

5. Find the slope at of the tangent to .
slope =

C. A curve is perpendicular to at . What is the slope of the tangent to at the point ? slope =

D. Give a formula for the slope at of the member of that goes through . The formula should not involve or .

E. A curve which at each of its points is perpendicular to the member of the family that goes through that point is called an orthogonal trajectory to . Each orthogonal trajectory to satisfies the differential equation
where is the answer to part D.
Find a function such that is the equation of the orthogonal trajectory to that passes through the point .