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Solution: APPM 1350 Exam 2 Spring 2011
- (21 pts) Each part is worth 7 points.
(a) Find the derivative of g(x) = cot(x^2 ) + cot^2 (x)
(b) Find an equation of the tangent line to the curve y =
|x| √ 2 − x^2
when x = 1.
(c) Find an equation of the tangent line to the curve x^2 + xy + y^2 = 3 at the point (1, 1).
Solution:
(a) f ′(x) = − csc^2 (x^2 ) · 2 x + 2 cot(x) · − csc^2 (x) = − 2 x csc^2 (x^2 ) − 2 cot(x) csc^2 (x)
(b) Note that if x > 0 then y =
x √ 2 − x^2
and y
′
(2 − x^2 )^3 /^2
, so y′(1) = 2 and y(1) = 1, so the
equation of the tangent line is y = 1 + 2(x − 1) = −1 + 2x
(c) Using implicit differentiation we have 2x + y + xy′^ + 2yy′^ = 0, so y′^ =
−(2x + y)
x + 2y
and y′(1, 1) = − 1
and so the equation of the tangent line is y = 1 + (−1)(x − 1) = 2 − x.
- (a) (10 pts) Find the linear approximation to f (x) =
25 − x^2 at x = 3.
(b) (10 pts) The radius of a circular disk is given as 4 cm with an error in measurement of ± 0 .2 cm.
Use differentials to determine the percentage error in the calculated area of the disk.
Solution:
(a) Note that f
′ (x) =
−x √ 25 − x^2
and L(x) = f (3) + f ′(3)(x − 3) = 4 − 34 (x − 3) = 254 − 34 x
(b) Note that A = πr^2 and so dA = 2πrdr and thus
dA
A
2 πrdr
πr^2
2 dr
r
so the relative error is ± 0 .1 and the percentage error is ±10%.
- (10 pts) Find the absolute maximum and absolute minimum values of f (x) = x
4 − x^2 on [− 1 , 2]
Solution:
Here f
′ (x) =
4 − 2 x^2 √ 4 − x^2
so f ′(x) = 0 when x = ±
2 and f ′(x) is undefined when x = ±2 so the
critical points are x = ±
The critical points in our interval are x =
2 , 2, and f (−1) = −
3, f (2) = 0 and f (
So the absolute minimum value is −
3 and the absolute maximum value is 2.
- (10 pts) A kite 30 ft above the ground moves horizontally at a speed of 10 ft/s. At what rate is the
angle between the kite-string and the ground decreasing when 50 ft of string has been let out?
Solution:
Let x be the horizontal distance the kite has traveled and let h be the length of string that has been
let out. Then we know dx/dt = 10ft/s. Now note that if θ is the angle that the string makes with
the ground then we have cot(θ) = x/30 and so
dθ
dt
= − sin^2 (θ) ·
dx
dt
rad/s
- (21 pts) Each part is worth 7 points.
(a) Find constants F and M such that the function y = F sin(x) + M cos(x) satisfies the differential
equation y′′^ + y′^ = sin(x).
(b) Suppose f (x) is differentiable for all values of x. Let B(x) = f (x) ◦ cos(x), find an expression
for B′′(x).
(c) Use the Mean Value Theorem on the function f (x) =
1 + x, for 0 ≤ x ≤ 1, to show that
Solution:
(a) Note that y′^ = F cos(x) − M sin(x) and y′′^ = −F sin(x) − M cos(x) and so,
y
′′
′ = (−F − M ) sin(x) + (F − M ) cos(x) = sin(x)
so −F − M = 1 and F − M = 0, and so F = M = −
(b) Here B(x) = f (cos(x)) and so, B′(x) = f ′(cos(x)) · − sin(x), and
B
′′ (x) = f
′′ (cos(x)) sin
2 (x) − f
′ (cos(x)) cos(x)
(c) Note that f ′(x) =
1 + x
and so by the Mean Value Theorem, there exists a c in (0, 1) such
that f (1) − f (0)
1 − 0
= f ′(c), that is,
1 + c
for some c in (0, 1)
and so
1 + c
and note that
1 + c
since
1 + c > 1 and so,
1 + c
and so we have
