If k is greater than or equal to 1, the graphs of y=sinx and y=ke^-x intersect for x is greater than or equal to 0.  Find the smallest value of k for which the graphs are tangent. k= ? What are...

If k is greater than or equal to 1, the graphs of y=sinx and y=ke^-x intersect for x is greater than or equal to 0.

 

Find the smallest value of k for which the graphs are tangent.
k= ?

What are the coordinates of the point of tangency?
x= ?, y= ? .

Asked on by bogshow24

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cosinusix | College Teacher | (Level 3) Assistant Educator

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If  the graphs are tangent at x it means that the graphs intersects and the slope of the tangent lines are the same.


i.e. `sinx=ke^(-x) ` and their derivatives are equal: c`os x=-ke^(-x)`

 

Therefore `sinx =ke^(-x)=-cos x`

 

Solve for x>0, sin x=-cos x

`pi/2+npi (n inZZ) ` is not a solution

Hence divide the inequality by `sin x!=0`

`tanx =-1`
` `

 

`x_n=3pi/4+n pi (n inNN)` Remember x>0

 

The solution of `sin x= -cos x` are `x_n=3pi/4+n pi (n inNN) `


The graphs intersects and are tangent if `sin x=-cos x` and `sinx =ke^(-x)`

Therefore `x=x_n ` for some n and `k=sin x_ne^x_n`

`sin x_n=+- sqrt(2)/2`

therefore `x=x_n` for some n and `k=+-sqrt(2)/2 e^(x_n)`

exp is an increasing function. The smallest |k| will occur for the smallest `x_n ` which is x_0 (n=0)

In this case `k=-sqrt(2)/2e^(3pi/4)`

 

Remark: I considered the smallest value of |k| because for n even, `sin x_n =-sqrt(2)/2` and  `k=-sqrt(2)/2 e^x_n)`

it is a decreasing negative function that goes to `-oo`. It doesn't have a smallest element but if we omit the sign, it does.

 

2nd question:

At the point of tangency, `x=x_0 =3pi/4 `

`y=sin x_0=-sqrt(2)/2`

 

Solution `k=-sqrt(2)/2 e^(3pi/4)`

`x= 3pi/4 `

`y=-sqrt(2)/2`

 

 

 

 

 

 

 

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