# Tutor profile: Patrick E.

## Questions

### Subject: Pre-Calculus

Give an example of polynomial of degree 3 with zeros at 3,-2, and 1.

The basic form of a polynomial is f(x)=a+b*x+c*x^2+d*x^3+...+z*x^n. However, there is an easier form that will allow us to answer the given problem. A polynomial can also be represented by its zeros, so for instance a polynomial with zeros (each of multiplicity 1) at A and B can be represented by f(x)=(x-A)*(x-B). So with our problem, we have zeros at 3,-2 and 1. So our polynomial will look like f(x) = (x-3)(x-(-2))(x-1) = (x-3)(x+2)(x-1). And there is our answer! But notice that if we multiply f(x) by any constant C, the above polynomial will still have zeros in all the right places. So f(x)=C(x-3)(x+2)(x-1) would also work for ANY number C. Isn't that interesting?

### Subject: Calculus

Is the sequence a_n=n/(n+1) convergent or divergent as n goes to infinity? Explain

If we naively take the limit as n goes to infinity, we see that the fraction is in indeterminant form (infinity / infinity) so we need to manipulate the fraction to see what the true limit is. If we divide both the numerator and denominator by the highest power (essentially multiplying the fraction by 1) we see that the fraction becomes (n/n) / ((n+1)/n) = 1 / (1+(1/n)) and if we let n go to infinity the numerator stays at 1 and the denominator goes to 1+0=1 (because 1/n goes to zero as n goes to infinity), so the entire fraction goes to 1/1=1. Since the limit of the sequence is a finite number, the sequence is convergent.

### Subject: Calculus

A particle is moving through the xy-plane with the given velocity and initial data. Find the position of the particle as a function of time. v(t)=sin(t)-cos(t), s(0)=0.

We know from the problem that the velocity of our particle is given by the function v(t)=sin(t)-cos(t), and from this we are looking for an equation for the position of the particle. We know that velocity is the derivative of position, i.e. s'(t)=v(t), so to find the position from velocity we will have to find the anti-derivative of v(t). The anti-derivative of sin(t) is -cos(t)+c1, and the anti-derivative of cos(t) is sin(t)+c2, where c1 and c2 are arbitrary constants. Therefore the anti-derivative of sin(t)-cos(t) is -cos(t)-sin(t)+C, where C=c1+c2. Therefore our general position equation is s(t)=-cos(t)-sin(t)+C. We're almost there! We now need to plug in our initial condition to find our constant C. Since we know s(0)=0, that must mean s(0)= -cos(0)-sin(0)+C = -1+C =0. Therefore C=1, so our position equation is given by s(t)=-cos(t)-sin(t)+1.

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