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What I understand is that where the t-value of $\frac{dy}{dt}$ and $\frac{dx}{dt}$ is equal to $0$, you can use that $t$ to determine the $x$ and $y$ coordinates of the point. What I got for my vertical tangent was $(e,1)$ and horizontal $(1,e)$ using a t-value of $t=1$ for vertical and $t=0$ for horizontal. This is wrong though.

F.A.
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dan
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  • One needs only one of the two derivatives to be 0 (not both at once, though if that happens it would also be solution). This is since question says "horizontal or vertical." – coffeemath Apr 22 '18 at 13:44

1 Answers1

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For horizontal tangents you want $$ \frac {dy}{dx}=0 $$and for vertical tangent lines you want $$ \frac {dx}{dy}=0$$

Using chain rule we get $$ \frac {dx}{dy} =\frac {dx/dt}{dy/dt} \\ \frac {dy}{dx} =\frac {dy/dt}{dx/dt}$$

Thus for horizontal tangent line we need $$\frac {dy}{dt}=0$$

Thus $$ -\sin te^{\cos t} =0 $$ which implies $t=0$ or $t=\pi$ The points are therefore, $(1,e)$ and $(1, 1/e)$

Similarly we find the vertical tangency happening at $(e,1)$ and $(1/e , 1).$

Mohammad Riazi-Kermani
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