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

[the7joker7]

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?

 

[Vid]

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.

 

[Nick R]

Hokay, what I did to solve this was find y as a function of $$\vartheta$$, then differentiate with respect to $$\vartheta$$, and set equal to 0, then solve for $$\vartheta$$.

so you have r = 6e^(0.4$$\vartheta$$)

and y = rsin($$\vartheta$$)

so y = 6sin($$\vartheta$$)e^(0.4$$\vartheta$$)

differentiating...

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

= 0

factor out the exponential term.

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

Now you have

6cos($$\vartheta$$) + 2.4sin($$\vartheta$$) = 0

since sin($$\vartheta$$) = 1 - cos^2($$\vartheta$$)

you get the quadratic

-2.4cos^2($$\vartheta$$) + 6cos($$\vartheta$$) + 2.4 = 0

solving you get

cos($$\vartheta$$) = -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 $$\vartheta$$ = 1.93 + N2$$\pi$$ 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?