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Finding a value of theta for which a tangent line is horizontal

  • Thread starter the7joker7
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  • #1
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Homework Statement



Find the smallest positive value for `theta` for which the tangent line to the curve `r = 6 e^(0.4 theta)` is horizontal

The Attempt at a Solution



I tried to use the (r'sin(theta)+rcos(theta))/(r'cos(theta) - rsin(theta)) formula to get the tangent line, which I got to be...

(6e^(0.4(theta))cos(theta))

divided by

-(6e^(0.4(theta))sin(theta))

And then find the theta that would make this equation 0, got pi/2, but the answer is 1.95130270391. Help?
 

Answers and Replies

  • #2
Vid
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Just a quick observation, but all of the 6e^(.4x)'s will cancel out leaving .4*sin(x)+cos(x) on top and .4cos(x) - sin(x) at the bottom.
 
  • #3
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Hokay, what I did to solve this was find y as a function of [tex]\vartheta[/tex], then differentiate with respect to [tex]\vartheta[/tex], and set equal to 0, then solve for [tex]\vartheta[/tex].

so you have r = 6e^(0.4[tex]\vartheta[/tex])

and y = rsin([tex]\vartheta[/tex])

so y = 6sin([tex]\vartheta[/tex])e^(0.4[tex]\vartheta[/tex])

differentiating...

dy/d[tex]\vartheta[/tex] = 6cos([tex]\vartheta[/tex])e^(0.4[tex]\vartheta[/tex]) + 2.4sin([tex]\vartheta[/tex])e^(0.4[tex]\vartheta[/tex])

= 0

factor out the exponential term.

The exponential term cannot be equal to 0 so divide it out.

Now you have

6cos([tex]\vartheta[/tex]) + 2.4sin([tex]\vartheta[/tex]) = 0

since sin([tex]\vartheta[/tex]) = 1 - cos^2([tex]\vartheta[/tex])

you get the quadratic

-2.4cos^2([tex]\vartheta[/tex]) + 6cos([tex]\vartheta[/tex]) + 2.4 = 0

solving you get

cos([tex]\vartheta[/tex]) = -0.3508 or 2.851

2.851 is not in the range of cos (wtf?)

-0.3508 is, so take the arccosine of that.

you get [tex]\vartheta[/tex] = 1.93 + N2[tex]\pi[/tex] where N is an integer

Although my answer is off by 2 hundredths of what you say it is.... how did you come up with 1.95?
 

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