2 Applications (do this after Matlab grader!)Let's use your root-finding code onan engineering application! A German mechanicalengineer named Bekker has studied the properties of off road vehicles. One of his resultsis that, for the illustration shown, the maximum angle at which a nose dive does notoccur is the solution of the equation:Asincos+Bsin2=Ccos+EsinwhereA=lsin1,B=lcos1,C=(h+0.5D)sin1-0.5Dtan1,E=(h+0.5D)cos1-0.5D.(a)(1 point) Convert this to a root-finding problem. You may leave your equation interms ofA,B,C, and E.(b)(2 points) With l=100 inches, h=40 inches, D=30 inches, and 1=10 degrees.By the picture, the angle is between 0 and 90 degrees. Use your secant.m witha tolerance 10-81e-8in MATLAB works in radians by default, but you know how to convert to degrees. Report theanswer only (no code), out to the 5th decimal place.(c)(2 points)We can check our answer by evaluating the function at the estimatedroot. Do this, and report your findings. How does this confirm that you have foundan approximate root? (Note: if this doesn't confirm that you've found a root, youshould reconsider your work.)After injecting A units of medicine at time t=0, the concentration of the medicine inthe bloodstream isc(t)=Ate-t4. The maximum safe concentration for this drug is1mgmL.(a)(3 points) How much medicine should be injected in order to reach the maximumsafe concentration? Use calculus (derivative, critical point, etc)to determine thetime at which c(t) reaches a maximum, then solve for the A value that will reachthis maximum. (Thisis just a pen and paper problem from Calculus; Matlab is notneeded.)(b)(4 points) Determine, using Newton's method, when the concentration falls to0.25mgmLc(t) equals 0.25at two times, but it only falls to0.25at one timef(x)=2x-x2100-30xx+1. Finding the roots of functions like thishas applications in molecular biology and population modeling.(a)(1 point) One of the roots isx*=0. Show (by-hand, not by plot) that x=0isaroot off. Show enough computation that it's clear to see what you did.(b)(3 points) One of the roots is between 150 and 200. Let's see which method can findthe root the fastest! Find an approximation to the root using all three methods.For bisection, use an initial interval of150,200. For secant method, use initialguesses ofx1=150 and x2=200. For Newton's method, use anin(d)(2 points) You should be surprised by the results. Especially with Newton's method.The reason why Newton's method isso slow is that our algorithm performed sym-bolic differentiation. This makes it nice on the user, but it ruins the speed ofNewton's method. Create an alternative algorithm for Newton's method. Struc-ture your code so that the function also inputs the derivative:1 function [p,n]= fastNewton