Math Questions?

Question

A curve has slope 2x+3 at each point (x,y) on the curve. Which of the following is an equation for this curve if it passes through (1,2)?

A function f has a Maclaurin series given by x^4/2! + x^5/3! + x^6/4! +...+ x^(n+3)/(n+1)!+... Write an expression for f(x).

The number of moose in a national park is modeled by the function M that satisfies the differential equation dM/dt = 0.6M(1-M/200) where t is the time in years and M(0)=50. What is the limit of M(t) as t approaches infinity?

What are all the values of p for which the infinite series summation of n / (n^p + 1) from n=1 to infinity converges?

Answer 1

#1

The slope at the point (1,2) is: 2(1)+3 = 5

Use point-slope form to find:

y - 2 = 5(x-1)

y = 5x - 3

#2

Factor out an x²

f(x) = x²(x²/2! + x³/3! + ... )

Now a bit of a trick (add and subtract the same quantities:

f(x) = x²(-1/0! - x/1! + 1/0! + x/1! + x²/2! + x³/3! + ... )

Take out the extra quantities:

f(x) = -x² - x³ + x²(1/0! + x/1! + x²/2! + x³/3! + ... )

That's the power series for e^x

f(x) = x²e^x - x² - x³

#3

The stabilizing population values are where of dM/dt is 0, which are at M=0 and 1-M/200 = 0, meaning M=200.

Between 0 and 200, in the viable range, dM/dt is positive, so it approaches M=200 as time goes on.

#4

p>2, although I'm not sure what kind of calculus you're supposed to use to figure this out. it depends on what kind of math course you're in.

you can sort of see that we require n / (n^p + 1) to shrink "faster" than 1/n (the harmonic series, the "borderline" case for a p-series), which by loose application of L'Hôpital's rule would allow us to say that p-1>1, meaning p>2.

Answer 2

A curve has slope 2x+3 at each point (x,y) on the curve. Which of the following is an equation for this curve if it passes through (1,2)?

slope=2*1+3=5

y-2=5(x-1)

y=5x+3 is the equation of the tangent line

Answer 3

y=x^2+3x+c and 2=1+3+c so c= -2
and y= x^2+3x-2
y= x^2( x^2/2!+x^3/!+++)= = x^2(e^x-1-x)
0.6 dM/M(200-M)/200)=dt
120dM/( A/M +B/(200-M) =dt
A(200-M) +B*M =1 so A= 1/200 and B=1/200
and 120/200(ln M-ln(200-M)) = t+C
(( M/(200-M))^3/5=Ke^t
K=(1/3)^3/5 so M/(200-M) =1/3 e^5/3*t
M = 200/3 e^5/3 t -1/3 e^5/3 t *M
so M= 200/3 e^5/3t/(1+1/3e^5/3t)
lim M t==> infinity = 200 ( please check calculations)
the series is of the same class as 1/n^(p-1) which converges for
p-1>1 so p>2