. Consider the fun ion defined implicitly near (2, 1) by x 2 y 2 + 3xy = 10y.
(a) [ points] Use implicit differentiation to find the derivative at (2, 1).
d dx
h x 2 y 2 + 3xy i
= d
dx [10y]
2xy 2 + 2yx 2 dy
dx + 3y + 3x dy
dx = 10 dy dx 2yx 2 dy
dx + 3x dy
dx − 10 dy
dx = −2xy 2 − 3y dy
dx = − 2xy 2 + 3y 2yx 2 + 3x − 10 Substituting x = 2 and y = 1, we get
dy dx
(2,1) = − 4 + 3 8 + 6 − 10 dy
dx
(2,1) = − 7 4
(b) [ points] Determine the equation of the tangent line at (2, 1).
The equation of the tangent line is y = − 7
4 (x − 2) + 1 .
. Let f (x) be the fun ion defined by f (x) = e √ x for x ≥ 0.
(a) [ points] Write an equation for the line tangent to the graph of f (x) at the point where x = 1.
f ′ (x) = d
dx e √ x = e √ x d dx
√ x = 1 2 √
x e √ x , So the slope of the tangent line at x = 1 is f ′ (1) = e
2 . The equation for the tangent line is y = e
2 (x − 1) + e .
(b) [ points] Find the point where this tangent line crosses the x-axis.
The line crosses the x-axis when y = 0.
e
2 (x − 1) + e = 0 e
2 (x − 1) = −e x − 1 = −2 x = −1 The tangent line crosses the x-axis at (−1,0).
. Let f be the fun ion defined by f (x) = x 3 − x 2 − 2x for 0 ≤ x ≤ 3.
(a) [ points] Find where the graph of f crosses the x-axis. Show your work.
Set f (x) = 0 and solve
x 3 − x 2 − 2x = 0 x(x 2 − x − 2) = 0 x(x + 1)(x − 2) = 0
The solutions are x = −1, x = 0, x = 2, but we are restri ed to [0,3], so the only two places where f crosses the x-axis will be x = 0 and x = 2.
(b) [ points] Find the intervals on which f is increasing.
Find the derivative and the critical points.
f ′ (x) = 3x 2 − 2x − 2 Now solve
3x 2 − 2x − 2 = 0 x = 2 ± √
28
6 = 1 ± √ 7 3 Only x = 1 + √
7
3 lies in our interval. We know that f (x) is decreasing if f ′ (x) < 0 which happens when 0 < x < 1 + √
7
3 . Also, f (x) is increasing if f ′ (x) > 0 which happens when 1 + √
7
3 < x < 3.
(c) [ points] Find the absolute maximum and absolute minimum value of f . Justify your answer.
f (x) is continuous on the closed interval [0, 3], so it attains its maximum and minimum. There are three points to check: x = 0, x = 1 + √
7
3 , and x = 3.
Spring MATH -
From the First Derivative Test we can see that the absolute minimum occurs at x = 1 + √
7
3 . Evaluating f at and , we have that f (0) = 0 and f (3) = 12, so the absolute maximum occurs at x = 3.
The maximum value of f on [0, 3] is and the minimum value is − 20 + 14 √ 7
27 ≈
−2.1126.
. [ points] A proje ile is fired toward the sea from a cliff feet above the water.
If the launch angle is 45 ◦ and the muzzle velocity is k, then the height h of the proje ile above the water when it is x feet downrange from the launch site is
h(x) = −32
k 2 x 2 + x + 200 feet.
See the figure.
Determine the maximum values of h and the distance downrange where this maxi- mum occurs if k = 2800 feet per second.
x
h
cliff
sea level
h(x) = − 32
k 2 x 2 + x + 200 h ′ (x) = − 64
k 2 x + 1 Set equal to zero and solve
− 64
k 2 x + 1 = 0 x = k 2
64
Since h ′′ (x) < 0, this will be the absolute maximum. Thus, the downrange distance is x = k 2
64 = 122, 500 feet. The height at that distance is h(122500) = 61, 450 feet
Spring MATH -
. Find the requested quantities in each of the following.
(a) [ points] Let
f (x) =
2 + cos(x − 1) if 0 ≤ x ≤ 1, ax − 2 if 1 < x < 2, x 2 − 2x + b if 2 ≤ x ≤ 3.
where a and b are constants. Find the values of a and b which make f (x) is continuous on [0, 3].
We need lim
x →1
−f (x) = lim
x →1
+f (x). This gives the equation 2 + cos(0) = a − 2
a = 5 Likewise, we need lim
x →2
−f (x) = lim
x →2
+f (x). This gives 2a − 2 = 2 2 − 2(2) + b
8 = 4 − 4 + b b = 8
These values for a and b will make f (x) continuous.
(b) [ points] Evaluate lim
n →∞
6n 2 − 3n + 7 (n + 3) 2 .
n lim →∞
6n 2 − 3n + 7
(n + 3) 2 = lim
n →∞
6n 2 − 3n + 7 n 2 + 6n + 9
= lim
n →∞
6 n 2 n 2 − 3 n
n 2 + 7 n 2 n 2
n 2 + 6 n n 2 + 9
n 2
= 6 1 = 6
. Find the following limits (a) [ points] lim
x →0
sin(x) x cos(x) .
Use l’Hospital’s Rule since both the numerator and denominator go to .
x lim →0
sin(x)
x cos(x) = lim
x →0
cos(x) cos(x) − x sin(x)
= 1
1 − 0 = 1
Spring MATH -
This can also be done as follows:
x lim →0
sin(x)
x cos(x) = lim
x →0
sin(x) x
1 cos(x)
= 1 · 1 = 1
(b) [ points] lim
x →∞
ln(xe x )
x .
Use l’Hospital’s Rule since both the numerator and denominator go to infin- ity.
x lim →∞
ln(xe x )
x = lim
x →∞
1
xe x (e x + xe x ) 1
= lim
x→∞
e x + xe x xe x
= lim
x →∞
1
x + 1 = 1
(c) [ points] lim
x →2
x 2 − 4 x 2 − x + 6 .
l’Hospital’s Rule does not apply.
x lim →2
x 2 − 4 x 2 − x + 6 = 0
8
= 0
. Find the following general antiderivatives.
(a) [ points]
Z
x 2 + x + x −1 + x −2 dx Z
x 2 + x + x −1 + x −2
dx = 1 3 x 3 + 1
2 x 2 + ln |x| − 1 x + C.
(b) [ points]
Z
2 sin(x) − 2cos(x) + 2sec 2 (x) dx Z
2 sin(x) − 2cos(x) + 2sec 2 (x)
dx = −2cos(x) − 2sin(x) + 2tan(x) + C.
(c) [ points]
Z e x − e −x 2
! dx.
Z e x − e −x 2
!
dx = 1 2
Z
(e x − e −x )dx = 1
2 (e x + e −x ) + C
Spring MATH -
. Use the Fundamental Theorem of Calculus (part ) to evaluate the following deriva- tives.
(a) [ points] If F(x) = Z x
1
cos 2 (u)e −u2du then F ′ (x) equals
By the Fundamental Theorem of Calculus, Part I, we know that F ′ (x) = d
dx Z x
1
cos 2 (u)e −u2du = cos 2 (x)e −x2.
.
(b) [ points] If G(x) = Z 3x2
1
cos(2u) sin(2u) du then G ′ (x) equals
By the Fundamental Theorem of Calculus, Part I, and the Chain Rule we have
G ′ (x) = d dx
Z 3x2
1
cos(2u) sin(2u) du
= cos
2(3x 2 ) sin
2(3x 2 ) d dx (3x 2 )
= 6x cos(6x 2 ) sin(6x 2 )
. Find the following definite integrals.
(a) [ points]
Z 3
1 4x − 1 x dx
Z 3
1 4x − 1
x dx = 2x 2 − ln(x)
3 1
= (18 − ln3) − (2 − ln1)
= 16 − ln3 ≈ 14.9014
(They may use the calculator to compute this integral, but in that case the answer is either right or wrong — no partial credit.)
Spring MATH -
(b) [ points] The fun ion h(x) is graphed below.
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6
Compute Z 10
0
h(x) dx.
This is to be done geometrically.
Z 10 0
h(x) dx = Z 1
0
h(x) dx + Z 2
1
h(x) dx + Z 4
2
h(x) dx + Z 5
4
h(x) dx + Z 10
5
h(x) dx
= (0)(1) + 1
2 (2 + 4)(1) + 1
2 (4 + 2)(2) + (2)(1) + 1
2 (1 + 5)(5)
= 26
. Find the requested areas.
(a) [ points] Set up (but do NOT evaluate) the integral that will give the area enclosed by the curves y = cos(x), y = 1+sin(2x) and the vertical line x = 3π
2 .
Π
Π
2
3 Π 2
