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Quiz 6 Solutions for MATH 161: Finding Tangent Lines and Applying Mean Value Theorem, Quizzes of Calculus

The solutions to quiz 6 of the math 161 course at millersville university. The quiz covers finding the equation of the tangent line to a curve and applying the mean value theorem. The steps for finding the slope of the tangent line and the equation of the tangent line using implicit differentiation and the point-slope formula.

Typology: Quizzes

Pre 2010

Uploaded on 08/18/2009

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Millersville University Name
Department of Mathematics
MATH 161, Quiz 6
February 27, 2004
Please answer the following questions. Your answers will be evaluated on their correctness,
completeness, and use of mathematical concepts we have covered. Please show all work and
write out your work neatly. Answers without supporting work will receive no credit.
1. Find the equation of the tangent line to the curve x3y3= 9yat the point with coordi-
nates (1,3). You may use implicit differentiation.
Applying implicit differentiation produces:
d
dx hx3y3i=d
dx [9y]
3x2y3+x3d
dx hy3i= 9y0(x)
3x2y3+x33y2y0(x) = 9y0(x)
3x2y3=3x3y2y0(x) + 9y0(x)
3x2y3=3x3y2+ 9y0(x)
x2y3=x3y2+ 3y0(x)
x2y3
3x3y2=y0(x)
Thus the slope of the tangent line at the point with coordinates (x, y) = (1,3) is
m=y0(1) = 1233
31332=27
39=27
6=9
2.
Using the point-slope formula for a line we get as the equation of the tangent line:
m=yy1
xx1
9
2=y3
x1
9
2(x1) = y3
9
2x+9
2=y3
9
2x+9
2+ 3 = y
9
2x+15
2=y
pf2

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Millersville University Name Department of Mathematics MATH 161, Quiz 6 February 27, 2004

Please answer the following questions. Your answers will be evaluated on their correctness, completeness, and use of mathematical concepts we have covered. Please show all work and write out your work neatly. Answers without supporting work will receive no credit.

  1. Find the equation of the tangent line to the curve x^3 y^3 = 9y at the point with coordi- nates (1, 3). You may use implicit differentiation. Applying implicit differentiation produces:

d dx

[ x^3 y^3

]

d dx

[9y]

3 x^2 y^3 + x^3

d dx

[ y^3

] = 9 y′(x)

3 x^2 y^3 + x^33 y^2 y′(x) = 9 y′(x) 3 x^2 y^3 = − 3 x^3 y^2 y′(x) + 9y′(x) 3 x^2 y^3 =

( − 3 x^3 y^2 + 9

) y′(x)

x^2 y^3 =

( −x^3 y^2 + 3

) y′(x) x^2 y^3 3 − x^3 y^2

= y′(x)

Thus the slope of the tangent line at the point with coordinates (x, y) = (1, 3) is

m = y′(1) =

Using the point-slope formula for a line we get as the equation of the tangent line:

m =

y − y 1 x − x 1

y − 3 x − 1 −

(x − 1) = y − 3

x +

= y − 3

x +

  • 3 = y

x +

= y

  1. Check the hypotheses of the Mean Value Theorem and find a value of c which makes the conclusion true for the function f (x) = x^3 + x^2 on the interval [0, 1]. The function f (x) = x^3 +x^2 is a polynomial and thus is continuous on the interval [0, 1] and differentiable on the interval (0, 1). Thus according to the Mean Value Theorem there exists a number 0 < c < 1 such that

f ′(c) =

f (1) − f (0) 1 − 0

13 + 1^2 − (0^3 + 0^2 )

Since f ′(x) = 3x^2 + 2x we need to solve the equation

3 c^2 + 2c = 2 3 c^2 + 2c − 2 = 0

c =

√ 22 − 4(3)(−2) 2(3)

=

Since −^1 −

√ 7 3 <^ 0 we will ignore this solution.^ The^ c^ promised by the Mean Value Theorem is c =

Note that 0 < c < 1.