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EXPERIMENT 1-A
1. Using MATLAB find the tangent to the curves y= √ x at x=4 at show graphically.
Solution:
The tangent to the line of the curve y=f(x) at the point P (a, f(a)) is the line through P with the slope m=f’(a)=
lim
h→ 0
f ( a + h ) − f ( a )
h
The following code illustrate the plotting of the tangent to the curve y= √ x at the point x=4.
clear all clc syms x y f = input (‘Enter the given function in variable x:’); x0 = input (‘Enter the x-coordinate of the point:’); y0 = subs (f, x, x0); fx = diff (f, x); m = subs (fx, x, x0); tangent = y0+m*(x-x0); t_line = y-tangent; plotrange = [x0-3, x0+3]; ezplot (f, plotrange); hold on; ezplot (tangent, plotrange) title (‘Tangent line plot’) t = sprintf (‘The tangent to the curve y=%s at (%d, %d) is y=%s’, f, x0, y0, tangent’); disp (t)
Input : Enter the given function in variable x: sqrt(x)
Enter the x-coordinate of the plot: 4
Output: The tangent to the curve y=x^ (1/2) at (4,2) is y=x/4 + 1
2. Using MATLAB find the tangent to the curves y=-sin(x/2) at the origin and show graphically.
Solution:
The tangent to the line of the curve y=f(x) at the point P (a, f(a)) is the line through P with the slope m=f’(a)=
lim h→ 0 f ( a + h ) − f ( a ) h
The following code illustrate the plotting of the tangent to the curve y=-sin(x/2)
clc syms x y f = input (‘Enter the given function in variable x:’); x0 = input (‘Enter the x-coordinate of the point:’); y0 = subs (f, x, x0); fx = diff (f, x); m = subs (fx, x, x0); tangent = y0+m*(x-x0); t_line = y-tangent; plotrange = [x0-3, x0+3]; ezplot (f, plotrange); hold on; ezplot (tangent, plotrange) title (‘Tangent line plot’) t = sprintf (‘The tangent to the curve y=%s at (%d, %d) is y=%s’, f, x0, y0, tangent’); disp (t)
Input : Enter the given function in variable x: -sin(x/2)
Enter the x-coordinate of the point : 0
Output : The tangent to the curve y= -sin(x/2) at (0,0) is y=-x/
3. Verify Rolle’s theorem for the function (^) ( x + 2 )^3 ( x − 3 )^4 in the interval [ 2,3]. Plot the curve along with the
secant joining the end points and the tangents at points which satisfy Rolle’s theorem.
Solution:
Rolle’s theorem: Suppose that the function y f (x) is continuous at every point of the closed interval [a, b] and
differentiable at every point in (a, b) and if f (a) f (b), then there is at least one number c in (a, b) at which f (c)
The below code illustrates the verification of Rolle’s theorem for the function f(x)= ( x + 2 )^3 ( x − 3 )^4 on the
interval [-2, 3].
4. Verify Lagrange’s mean value theorem for the function f(x)=x+e^3 x^ in the interval [0,1]. Plot the curve along with
the secant joining the end points and the tangents at points which satisfy Lagrange’s mean value theorem.
Solution:
Suppose that the function y f (x) is continuous at every point of the closed interval [a, b] and differentiable at
every point in (a, b), then there is at least one number c in (a, b) so that
f’(c)= f ( b )− f ( a ) b − a The below code illustrates the verification of Lagrange’s theorem for the function f(x)=x+e^3 x^ on the interval [0,1] clear all; clc; syms x c; f=input ('Enter the function: '); I=input ('Enter the interval [a, b]: '); a=I (1); b=I (2); fa=subs (f, x, a); fb=subs (f, x, b); df=diff (f, x); dfc=subs (df, x, c); LM=dfc-(fb-fa)/(b-a); c=solve (LM); count=0; for i=1: length(c) if c(i)>a && c(i) count=count+1; tx (count)=c(i); end end fprintf ('The values of c between %d and %d which satisfy LMVT are x=', a, b); disp(double(tx)) xval=linspace (a, b,100); yval=subs (f, x, xval); m=subs (df, tx); % Slopes of tangents at the points between a and b. ty=subs (f, x, tx); plot (xval, yval); hold on;
secant_slope=(fb-fa)/(b-a); secant_line=fa+secant_slope(x-a); sx_val=xval; sy_val=subs (secant_line, x, sx_val); plot (sx_val, sy_val); for i=1: length(tx) txval=linspace(tx(i)-1, tx(i)+1,20); t_line=ty(i)+m(i)(x-tx(i)); tyval=subs (t_line, x, txval); plot (txval, tyval,'k'); hold on plot(tx(i), ty(i),'ok'); end hold off; grid on legend ('Function', 'Secant', 'Tangents'); title ('Demonstration of LMVT');
Input : Enter the function: x+exp(3*x)
Enter the interval [a, b]: [0,1]
Output : The values of c between 0 and 1 which satisfy LMVT are x=0.
