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MATH 1220 PRACTICE MIDTERM 3 SOLUTIONS
MIKE WILLS
- Problem 1 Show that the point given in polar coordinates by
2 , π 2
lies on the polar curves given by
4 = r^2 sin θ r = 2 cos(2θ).
Solution 1. Let P be the point in question. Note that
(2) P =
π 2
3 π 2
Since
(3) 22 sin
(π 2
P lies on the first curve. Since
(4) −2 cos
3 π 2
P lies on the second curve.
- Problem 2 Find the equation of the tangent line at t = 1 to the curve given parameterically by
x(t) = sin(πt) y(t) = et 2 .
Solution 2. We compute:
x(1) = sin π = 0 y(1) = e^1 2 = e x ˙(t) = π cos(πt) y˙(t) = 2tet^2 x ˙(1) = π cos(π) = −π y ˙(1) = 2(1)e^12 = 2e.
The slope of the tangent line is
(7) y˙(1) x˙(1)
2 e π
The equation of the tangent line in point-slope form is therefore
(8) y − e = − 2 ex π
or equivalently,
(9) y = e − 2 ex π
- Problem 3 Show that the sequence
(10)
{(−1)n n
n=
converges to zero.
Solution 3. Let an = (−1) n n. Notice that
(11) −
n ≤ an ≤
n
Since
(12) (^) nlim→∞
n
it follows by the squeeze theorem that
(13) (^) nlim→∞ an = 0.
- Problem 6 Compute the radius of convergence of the series
(19)
∑^ ∞
n=
xn n^2
Solution 6. Write an = (^) n^12. The radius of convergence is given by
(20) R = lim n→∞
an an+
∣ = lim n→∞
(n + 1)^2 n^2
