15. (1 pt) Consider the curves in the first quadrant that have equations y- Aexp
ID: 2888137 • Letter: 1
Question
15. (1 pt) Consider the curves in the first quadrant that have equations y- Aexp(2x), where A is a positive constant. Different values of A give different curves. The curves form a family, F. Let P (6,2). 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 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(v) 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 =- , g(v) where g() is the answer to part D. Find a function h(v) such that x h(y) is the equation of the orthogonal trajectory to F that passes through the point P.Explanation / Answer
15)
y=Ae2x passes through P(6,2)
A)
at point (6,2)
2=Ae2*6
=>2=Ae12
=>A=2e-12
f(x)=2e-12e2x
=>f(x)=2e2x-12
B)
f '(x)=2e2x-12(2*1 -0)
=>f '(x)=4e2x-12
at x=6, slope of tangent to C=f '(6)
=>slope of tangent to C=4e2*6-12
=>slope of tangent to C=4e0
=>slope of tangent to C=4*1
=>slope of tangent to C=4
C)
product of slopes of perpendicular lines =-1
=>slope of tangent to C*slope of tangent to D =-1
=>4*slope of tangent to D =-1
=>slope of tangent to D = -1/4
D)
f(x)=Ae2x
=>f '(x)=Ae2x *2*1
=>f '(x)=2y
=>g(y)=2y
E)
dy/dx =-1/2y
=>y dy =-(1/2)dx
=> y dy = -(1/2)dx
=>(1/2)y2=-(1/2)x +c
=>y2=-x +c
=>x=c-y2
curve passes through (6,2)
=>6=c-22
=>6=c-4
=>c=10
=>x=10-y2
=>h(y)=10-y2
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