2 x dy dx 2 x 2 y x y solution ii solve for y and

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Precalculus: Mathematics for Calculus
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Chapter 3 / Exercise 61
Precalculus: Mathematics for Calculus
Redlin/Stewart
Expert Verified
2 x = dy dx = 2 x 2 y = x y
Solution ii. Solve for y and then di ff erentiate using the Chain Rule. First, x 2 + y 2 = 4 = y 2 = 4 x 2 = y = ±
(4 x 2 ) . Second, dy dx = d dx ±
(4 x 2 ) = 1 ± 2
(4 x 2 ) · d dx
4 x 2
= 1 ± 2
(4 x 2 ) · (0 2 x ) = 2 x ± 2
(4 x 2 ) = x ±
(4 x 2 ) = x y .
c. y =
x 0 t cos(3 t ) dt Solution. By the Fundamental Theorem of Calculus: dy dx = d dx
x 0 t cos(3 t ) dt = x cos(3 t )
d. y = ln
x 3
Solution i. Simplify, then di ff erentiate. First, y = ln
x 3
= 3ln( x ). Second, dy dx = d dx 3ln( x ) = 3 · 1 x = 3 x .
2 187
We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
Precalculus: Mathematics for Calculus
The document you are viewing contains questions related to this textbook.
Chapter 3 / Exercise 61
Precalculus: Mathematics for Calculus
Redlin/Stewart
Expert Verified
Solution ii. Di ff erentiate using the Chain Rule, then simplify: dy dx = d dx ln
x 3
= 1 x 3 · d dx x 3 = 1 x 3 · 3 x 2 = 3 x
3. Do any two (2) of a c . [10 = 2 × 5 each] a. Explain why lim x 0 x | x | doesn’t exist. Solution. Note that when x > 0, | x | = x , so x | x | = 1, and when x < 0, x = | x | , so x | x | = 1. It follows that lim x 0 x | x | = lim x 0 1 = 1 and lim x 0 + x | x | = lim x 0 + 1 = 1, so lim x 0 x | x | can’t exist since 1
= 1.
b. A spherical balloon is being inflated at a rate of 1 m 3 /s . How is its radius changing at the instant that it is equal to 2 m ? [The volume of a sphere of radius r is V = 4 3 π r 3 .] Solution. On the one hand, we are given that dV dt = 1; on the other hand, using the Chain Rule, dV dt = d dt 4 3 π r 3 = 4 3 π
d dr r 3
dr dt = 4 3 π 3 r 2 dr dt = 4 π r 2 dr dt . It follows that 1 = dV dt = 4 π r 2 dr dt , so dr dt = 1 4 π r 2 . Thus, at the instant that r = 2 m , we have dr dt = 1 4 π 2 2 = 1 16 π m/s .
c. Use the Left-Hand Rule to find
3 1 x dx .
n 1
i =0 i = 0 + 1 + · · · + ( n 1) = n ( n 1) 2
Solution. Not letting the right hand know what the left hand is doing:
3 1 x dx = lim n →∞ n 1
i =0 3 1 n ·
1 + i 3 1 n
[Since our function is just f ( x ) = x .] = lim n →∞ n 1
i =0 2 n
1 + i 2 n
= lim n →∞ 2 n n 1
i =0
1 + i 2 n
= lim n →∞ 2 n
n 1
i =0 1
+
n 1
i =0 i 2 n
= lim n →∞ 2 n
n +
2 n n 1
i =0 i
= lim n →∞ 2 n
n + 2 n · n ( n 1) 2 i
= lim n →∞ 2 n ( n + ( n 1)) = lim n →∞ 2 n (2 n 1) = lim n →∞
4 2 n
= 4 0 = 4
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4. Let f ( x ) = x 2 x 2 + 1 . Find the domain and all the intercepts, vertical and horizon- tal asymptotes, and maxima and minima of f ( x ), and sketch its graph using this information. [11] Solution. We run through the checklist: i. Domain. f ( x ) = x 2 x 2 + 1 always makes sense because the denominator x 2 + 1 1 > 0 for all x . Thus the domain of f ( x ) is all of R ; note that f ( x ) must also be continuous everywhere.
ii. Intercepts. f (0) = 0, so (0 , 0) is the y -intercept. Since f ( x ) = x 2 x 2 + 1 = 0 is only possible when the numerator is 0, any x -intercepts occur when x 2 = 0, i.e. when x = 0. Thus (0 , 0) is the only x -intercept, as well as the y -intercept. <