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I need to find the equation of the tangent line to the point (1,0) for the equation:

$x=e^{-0.1t}cos(t) \\ y=e^{-0.1t}sin(t)$

I also need to calculate the area in the first quadrant bounded on the outside by the curve for $t$ greater than or equal to $0$ and less than or equal to $\pi/2$.

I am not totally sure what to do for either one.

I know that $\frac{dy}{dx}$ is $\frac{dy}{dt}/\frac{dx}{dt}$.

Is the tangent line equation $y=\frac{dy}{dx}(x-1)$? Once I calculate $\frac{dy}{dx}$ I'd just plug it in to the equation.

FofX
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1 Answers1

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Hint:

For the first part, you can find the (unnormalized) tangent vector $T(t)$ at the point $p(t) = (1,0)$ on the curve by calculating the derivative with respect to the parameter, $t$: $$T(t) = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}$$

For the second part, find the polar form of the curve $r = f(\theta)$. Then the area is $$\iint f(\theta) \ r \ dr \ d\theta$$ You'll also need to figure out the appropriate limits of integration.

Simon S
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  • Thanks for the help, but we haven't done vectors or double integrals so I don't believe this is the method I am supposed to use. – FofX Nov 25 '14 at 19:35
  • I guess then use your original idea of calculating $dy/dx$ as the ratio of $dy/dt$ and $dx/dt$. Not sure how you're going to get at the area without some integration. – Simon S Nov 25 '14 at 19:41
  • I'll edit my original post for the tangent line equation, I think I might have figured it out. And yes you are right I will need to integrate I'm just not sure how to set it up. Although I know that we didn't use a double integral. Is there another way you know of with a single integral? – FofX Nov 25 '14 at 19:48