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The solutions to exam ii for calculus i, taught by prof. P. Wong, on november 12, 2004. It includes problems on finding derivatives, estimating limits using local linearization and l'hopital's rule, and optimizing the area of a rectangle. Students can use this document as a reference for understanding the concepts and techniques covered in the exam.
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EXAM II - NOVEMBER 12, 2004
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
1
2 EXAM II - NOVEMBER 12, 2004
dy dx = 1^ ·^ cos(
√xex) + x · (− sin(√xex)) · d dx (
√xex)
= cos(√xex) − x sin(√xex) · (^12 exx−^1 /^2 + √xex).
(7 pts.)(b) y = (tan 2 + tan x)e.
dy dx =^ e^ ·^ (tan 2 + tan^ x)
e− (^1) · d dx (tan 2 + tan^ x) = e · (tan 2 + tan x)e−^1 · sec^2 x.
(7 pts.)(c) y = arcsin( (^) x^2 ).
dy dx =^
1 − ( (^2) x )^2
· (− (^) x^22 ).
4 EXAM II - NOVEMBER 12, 2004
Using the technique of implicit differentiation with respect to x, we have 2 x + (y + x dy dx ) − 2 y dy dx = 0
or 2 x + y + (x − 2 y) dy dx = 0.
It follows that dy dx =
2 x + y 2 y − x.
(9 pts.)(b) Find an equation for the line tangent to the graph of y at the point (2, 3).
At the point (2, 3), the slope of the tangent is given by dy dx
(2,3)
The tangent line is given by
(y − 3) =^74 (x^ −^ 2)
or y =^74 x − 12.
MATH105A CALCULUS I - PROF. P. WONG 5 4.(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. Draw a picture first!] Find the dimensions of the rectangle which has the maximum area.
Consider the picture below.
(x,y)
y=9−x 2
The upper right hand corner of the inscribed rectangle has co- ordinates (x, y) = (x, 9 − x^2 ). The area of this rectangle is given by A = 2xy since the base has length 2 x. Now,
A = 2xy = 2x(9 − x^2 ) = 18x − 2 x^3.
Taking the derivative of A with respect to x yields dA dx = 18^ −^6 x
By setting dA dx = 0, we obtain x = ±
3 and so y = 9 − (
3)^2 = 6. Thus, the dimensions of the maximized rectangle are 2
3 by 6.
