Hello, welcome to this part of the exercise on calculating a displacement at the surface of Earth. We will consider our terrestrial sphere with a tiny infinitesimal displacement at a geographical coordinate on Earth. We recall here the terrestrial sphere, that I draw here with its equator, the prime meridian, the rotational axis <i>z</i>, the axis that passes through the prime meridian <i>x</i>, and then one point, in coordinates, along a meridian, here, at a certain height <i>P</i>, I have here the latitude <i>φ</i>, and the longitude <i>λ</i>. What interests us in this exercise, is a tiny displacement, here, at the surface of Earth, at point <i>P</i>. If we now decompose its coordinates at the point <i>P</i>, we see that the point <i>P</i> is situated at a height, or latitude <i>φ</i>, and that, at this height, we will consider the radius of the parallel, here <i>r</i>, which is the radian of the parallel at the latitude <i>φ</i>. What interests us initially, is a small displacement, here, <i>dφ</i>, at this point i>P</i> which will generate here a displacement at the surface of Earth <i>R.dφ</i>, with <i>dφ</i> expressed in radians. By analogue, we can consider a small displacement in longitude, here <i>dλ</i>, which will result in my parallel, here, a displacement <i>r.dλ</i> also expressed in radians. Finally, what interests us is the component <i>d2</i>, that is to say, my <i>da</i> here, given that I will consider this <i>da</i> as a tiny displacement, and I can say that the triangle here is a rectangular triangle, with my component <i>R.dφ</i> on one side, and then <i>r.dλ</i> on the other. Firstly, we need to calculate the radius of the parallel, the <i>r</i>, radius of the parallel. We bring back here our latitude <i>φ</i>, and this <i>r</i>, is nothing else than the radius of the terrestrial sphere times the cosinus <i>φ</i>. If <i>φ</i> is zero, so that at the equator, <i>r</i> obviously equals <i>R</i> and if <i>φ</i> is equal to 90º, so at the poles, my <i>r</i> is equal to zero. I will now calculate my displacement, here, <i>da</i>, if I look at my triangle here, which I consider as a rectangle, the <i>da²</i> is nothing else than <i>(r.dλ)²</i> plus <i>(R.dφ)²</i> which is equal to <i>R²</i> times <i>cos²φ.dλ²</i> plus <i>R².dφ²</i>. We can highlight <i>R²</i> that multiply <i>cos²φ.dλ²</i> plus <i>dφ²</i>. So for this value <i>da²</i>, obviously, what interests us, is the root, that is <i>da</i>. I summarize here the approach, in plain text the formulas which are returned on this part of the screen. We take now a small numerical example, with a point <i>P</i>, located in Cameroon, at a latitude near the equator, or 4º, and a longitude of about 9º. We consider the infinitesimal displacement for one second, that means <i>1/3600</i> degrees. We recall here the radius of Earth, which is <i>6400 Km</i>. So the radius squared, expressed in meters, will be <i>4,096.10¹³</i> m. Our angular displacement, so <i>dφ</i> or <i>dλ</i>, is equal to one second, so we saw that it was <i>1/3600</i>, which equals to <i>2,77.10⁻⁴</i> degrees, which we will express and convert to radians, which equals to <i>4,85.10⁻⁶</i> radians. Taking our latitude, here at <i>4,03</i> degrees, I calculate the cosine of the latitude, namely <i>0,99753</i>. And my <i>da²</i>, if I take the formula seen a while ago, equal to <i>R²</i> times <i>cos²φ.dλ²</i> plus <i>dφ²</i>. Therefore, this, by calculating, I get <i>4,689.10⁻¹¹</i>, and this is equal to <i>1920,7</i>. Taking the root, we get a <i>da</i> of <i>43,8</i> meters. We give here, clearly again, a resolution of this equation, with here rounding to 44 meters for this numerical example.