How to draw a tangent line to the curve?(tangent line has to pass through origin)

Hi everbody, I plotted this particular curve,but ı have to draw a tangent line,which has to pass through origin, to the curve. I'd really appreciate your help.

Cd0 = 0.017;
Cl0 = 0.1;
K = 0.075;

Cl = 0:0.02:1.6;
Cd = Cd0 + K*(Cl - Cl0).^2;
plot(Cd,Cl)
    

This is how the plot looks like

[darova]

Is this correct?

[Furkan Yürümez]

first of all, thanks for your help but the line has to pass through origin(0,0)

[William Andriamihaja]

How did you make your drag polar look so smooth? I'm doing the same thing but mine looks horrible.

Answer 1

[John D'Errico]

You are plotting a parabola on its side. Then you want to find the point where a line that passes through the origin, is also tangent to the curve.

First, what is the equation of a line that passes through the origin, given that Cl and Cd are your variables. Here, the variable Cd lies on the x axis, so it is the independent variable. The only unknown is the slope, since the line must pass through the origin. So we would have a line with equation

Cl = s*Cd

for some unknown slope s. If the line intersects the curve at a tangent point, then there is only ONE point of intersection. Just solve the system of two equations. That is most easily done by eliminating one of the variables.

Cd = Cd0 + K*(s*Cd - Cl0).^2

If we expand this into a quadratic in a standard form, then look at what is inside the sqrt. I'll do this using MATLAB, so I can claim to have used MATLAB to solve a problem that really is just algebra.

syms Cd Cd0 Cl0 K s 
>> solve( Cd == Cd0 + K*(s*Cd - Cl0).^2,Cd)
ans =
 ((- 4*Cd0*K*s^2 + 4*Cl0*K*s + 1)^(1/2) + 2*Cl0*K*s + 1)/(2*K*s^2)
 (2*Cl0*K*s - (- 4*Cd0*K*s^2 + 4*Cl0*K*s + 1)^(1/2) + 1)/(2*K*s^2)

For a unique intersection to exist, that discriminant must be zero. So now we have:

- 4*Cd0*K*s^2 + 4*Cl0*K*s + 1 == 0

This allows us to compute the two possible slopes of lines that touch the parabola at a point of tangency...

solve(- 4*Cd0*K*s^2 + 4*Cl0*K*s + 1 == 0,s)
ans =
 (Cl0*K - (K*(K*Cl0^2 + Cd0))^(1/2))/(2*Cd0*K)
 (Cl0*K + (K*(K*Cl0^2 + Cd0))^(1/2))/(2*Cd0*K)

You want the positive slope of those two, which must be the latter. Therefore we have a line of tangency as:

Cl = s*Cd

where s is given as

s = (Cl0*K + (K*(K*Cl0^2 + Cd0))^(1/2))/(2*Cd0*K)

Just pick two points on the line, and connect them. With actual numbers, we would have...

Cd0 = 0.017;
Cl0 = 0.1;
K = 0.075;
Cl = 0:0.02:1.6;
Cd = Cd0 + K*(Cl - Cl0).^2;

s = (Cl0*K + (K*(K*Cl0^2 + Cd0))^(1/2))/(2*Cd0*K);

Clmax = 1.6;

plot(Cd,Cl,'b-',[0,Clmax/s],[0,Clmax],'r-')

Yes, there are other ways this could have been done, perhaps using the dreaded calculus. EEK! ;-) But this solution requires nothing more than understanding the quadratic formula, and a little algebra.

Answer 2

[Sulaymon Eshkabilov]

One quich-n-easy solution graphically is to use ginput() by which you can select two or more points and then plot the selected data.

Answer 3

[darova]

Here is the success condition:

Numerical approach

dx = 0.1;
f1 = @(x0) (interp1(x,y,x0+dx) - interp1(x,y,x0))/dx;
f2 = @(x0) (interp1(x,y,x0)-0)/(x0-0);

Or you can take derivative from your function