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

Question

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

Faiza Fuller · Accepted Answer

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Jeffrey Jordon · Accepted Answer

Consider the curve and point,  y=Ae7x;P(6,6)  a) Find the value of f(x) as follows:  The curve passes through the point P(6,6).  That implies,  6=Ae7(6)  A=6e42  Substitute A=6e42 in the curve y=Ae7x.  y=6e42e7x  Thus, the required value is f(x)=6e42e7x  b) Find the slope at P of tangent line C as follows:  Slope of the tangent line is,  dydx=ddx[6e42e7x]  =6e42ddx[e7x]  =6e42(7e7x)  =42e42e7x  At the point P(6,6),  dydx|(6,6)=42e42e7(6)  =42e42e42  =42  Thus, the slope of the tangent line at the point P(6,6) is,  dydx|(6,6)=42  c) Find the slope of the curve at the curve D perpendicular to C at the point P.  As D is perpendicular, slope of D is,  =1dy/dx  =−142  Thus, the required slope is −142  d) From the above part (b),  dydx=42e42e7x  =6⋅7e42e7x  =7(6e42e7x)  =7y  Therefore, the slope of the tangent line in terms of y is,  dydx=7y  e) Find the function x=h(y) as follows:  Consider,  dydx=−1g(y)  From the above part (d),g(y)=7y  Substitute g(y)=7y in the differential equation dydx=−1g(y)  dydx=−17y  Apply the separation of variables,  −7ydy=dx  Integrate on the both sides,  ∫−7ydy=∫dx  −7(y22)+c=x  x=−7y22+c  h(y)=−7y22+c  Thus, the required values is,  h(y)=−7y22+c

Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. T

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