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This is the Solved Exam of Calculus which includes Rambling, Worst Three, Rabbits, Commonly, Proportional, Average Rate, Proper Justiffication, Derivative etc. Key important points are: Property, Function, Derivative, Curve, Equation, Line Tangent, Curve, Point, Limits, Algebraically
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EXAM II - NOVEMBER 12, 2010
Instruction: Read each question carefully. Explain ALL your work and give reasons to support your answers.
Advice: DON’T spend too much time on a single problem.
Problems Maximum Score Your Score
Total 100
1
1.(15 pts.) Let r(x) be a function with the property that r(0) = 2 and r′(0) = −1.
(i) Suppose f (x) = r(x) sin x. Find f ′(0).
Using the product rule, we have
f ′(x) = r′(x) sin x + r(x) cos x f ′(0) = r′(0) sin 0 + r(0) cos 0 = r(0) cos 0 = 2.
(ii) Suppose g(x) =
ex r(x)
. Find g′(0).
Using the quotient rule, we have
g′(x) = exr(x) − exr′(x) [r(x)]^2
g′(0) = e^0 r(0) − e^0 r′(0) [r(0)]^2
=
(iii) Suppose h(x) = ln(r(x)). Find h′(0).
Using the chain rule, we have
r′(x) =
r(x) · r′(x)
h′(0) = r′(0) r(0) = −
3.(14 pts.) Consider the curve given by y^2 − 6 xy + 20 = 0.
(i) Find
dy dx
Differentiating y^2 − 6 xy + 20 = 0 implicitly with respect to x, we have
2 y dy dx
y + x dy dx
Solving for (^) dxdy , we get dy dx
3 y y − 3 x
(ii) Find an equation of the line tangent to the curve at the point (2, 2).
Using the result from part (i), the slope of the desired tangent line at the point (2, 2) is given by dy dx
(2,2)
An equation of the tangent line is y − 2 x − 2
or 3 x + 2y = 10.
4.(18 pts.) Evaluate each of the following limits (if it exists) algebraically. (Do not evaluate the limits using calculator.) Explain your work.
(i) lim t→∞
3 t^2 + t + 1 7 − 5 t − 7 t^2
The limit is of type
. Thus, we can apply L’Hˆopital’s rule and we have
t^ lim→∞
3 t^2 + t + 1 7 − 5 t − 7 t^2
L’H = lim t→∞
6 t + 1 − 5 − 14 t (again of type
L’H = lim t→∞
(ii) lim x→ 1 2 x^ − 2 x − 1
The limit is of type
. Thus, we can apply L’Hˆopital’s rule and we have
x^ lim→ 1
2 x^ − 2 x − 1
L’H = lim x→ 1
(ln 2)2x 1 = 2 ln 2 = ln 4.
(iii) lim θ→ 0
3(1 − cos θ) 2 θ^2
The limit is of type
. Thus, we can apply L’Hˆopital’s rule and we have
lim θ→ 0
3(1 − cos θ) 2 θ^2
L’H = lim θ→ 0
3 sin θ 4 θ (again of type
L’H = lim θ→ 0
3 cos θ 4
6.(20 pts.) A rectangle is placed inside the region bounded by the parabola y = 9 − x^2 and the x-axis. [You may assume the bottom of the rectangle is on the x-axis and the upper corners of the rectangle are on the parabola.]
0
A = (x,y)
(a) Suppose the corner (see figure) A has coordinates (x, y). Express the area of the rectangle in terms of x and y.
The area R of such a rectangle is given by R = 2xy.
(b) Find the dimensions of the rectangle which has the maximum area.
To maximize R, we first write R as a function of one variable using the fact that the point A is on the parabola so that y = 9 − x^2. It follows that R = 2x(9 − x^2 ) = 18 x − 2 x^3. Now, dR dx = 18 − 6 x^2 set = 0
⇒x^2 = 3 or x = ±
When x = ±
3 , y = 9 − 3 = 6.
Thus, the dimensions of the rectangle are 2
3 by 6. To ensure that these dimen- sions yield maximum area, we can take the second derivative of R and get − 12 x so that when x =
3 , R′′^ = − 12 < 0 and thus the critical point x =
3 is a local max.
