i. We were given the parametric equations,
[tex]x=e^{-t}, y=e^{-t}sin(t)[/tex]
To find the equation of the normal, we need the gradient of the tangent to the cure at P. Then we take the negative reciprocal of it. So let us differentiate to get,
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
[tex]\frac{dy}{dx} =\frac{e^{-t}(cos(t)-sin(t)}{-e^{-t}}[/tex]
[tex]\frac{dy}{dx}=-(cos(t)-sin(t))[/tex]
We now substitute
[tex]t=\frac{\pi}{2}[/tex]
[tex]\frac{dy}{dx}=-(cos(\frac{\pi}{2})-sin(\frac{\pi}{2}))[/tex]
[tex]\frac{dy}{dx}=-(0-1)[/tex]
[tex]\frac{dy}{dx}=-(-1)[/tex]
[tex]\frac{dy}{dx}=1[/tex]
Hence the slope of the normal,
[tex]m=\frac{-1}{1}=-1[/tex]
We need one more thing to find the equation of this straight line. Do you remember what that is?
Good! We need a point, so we substitute,
[tex]t=\frac{\pi}{2}[/tex]
into
[tex]x=e^{-t}, y=e^{-t}sin(t)[/tex]
to get,
[tex]x=e^{-\frac{\pi}{2}}, y=e^{-\frac{\pi}{2}}sin(\frac{\pi}{2})[/tex]
This gives us,
[tex]x=e^{-\frac{\pi}{2}}, y=e^{-\frac{\pi}{2}}[/tex]
We now use the formula,
[tex]y-y_1=m(x-x_1)[/tex]
to get the equation of the normal.
We substitute to obtain,
[tex]y-e^{-\frac{\pi}{2}}=-1(x-e^{-\frac{\pi}{2}})[/tex]
This simplifies to,
[tex]y=-x+e^{-\frac{\pi}{2}}+e^{-\frac{\pi}{2}}[/tex]
[tex]y=-x+2e^{-\frac{\pi}{2}}[/tex]
iii) From the diagram(Please check second attachment), the normal is above the curve, C on the interval,
[tex]0\le t\le \pi[/tex]
So the area,
[tex]A=\int\limits^\pi_0 {y(t)} \, dt -\int\limits^\pi_0 {y(t)\times x'(t)} \, dt[/tex]
Meaning that we have to plug in the parametric forms of the equations.
This implies that
[tex]A=\int\limits^\pi_0{-e^{-t}+2e^{-\frac{\pi}{2}}\,dt -\int\limits^\pi_0 {e^{-t}sin(t)\times -e^{-t}\,dt[/tex]
This simplifies to;
[tex]A=\int\limits^\pi_0 {-e^{-t}+2e^{-\frac{\pi}{2}}\, dt -\int\limits^\pi_0 {-e^{-2t}sin(t) \, dt[/tex]
Now we apply integration by parts(which was given to you already) to evaluate the first integral and then integrate the second integral directly.
[tex]A=(e^{-t}+2e^{-\frac{\pi}{2}})|_0^{\pi}--\frac{2}{5}e^{-2t}sin(t)-\frac{1}{5}e^{-2t} |_0^{\pi}[/tex]
Evaluating this gives us,
[tex]A=2\pi e^{-\frac{\pi}{2}}+e^{-\pi}-1+\frac{e^{-2 \pi}+1}{5}[/tex]
iii) To find other points of intersections of the normal and the curve, C.
Solve both eqautions simultaneously.
We know the equation of the normal is,
[tex]y=-x+2e^{-\frac{\pi}{2}}[/tex]
and the curve C has equations defined parametrically by,
[tex]x=e^{-t}, y=e^{-t}sin(t)[/tex]
Subsitute x and y into the normal and solve to obtain,
[tex]e^{-t}sin(t)=-e^{-t}+2e^{-\frac{\pi}{2}}[/tex]
Solving for t gives,
[tex]t=-1.01, t=1.57[/tex]
