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Math 105: Review for Exam II - Solutions
- Find dy/dxdy/dxdy/dx for each of the following.
(a) yyy === xxx^222 + 2+ 2+ 2xxx^ +++ eee^222 +++ eee^222 xxx^ + ln 2 + ln (2+ ln 2 + ln (2+ ln 2 + ln (2xxx) + arctan 2) + arctan 2) + arctan 2 dy dx = 2x + (ln 2)2x^ + 2e^2 x^ +
2 x · 2 Note that e^2 , ln 2, and arctan 2 are constants.
(b) y =
y = x · arctan(5x)
y = x · arctan(5x)
x · arctan(5x) dy dx
x−^1 /^2 arctan(5x) +
x ·
1 + (5x)^2
arctan(5x) 2 x^1 /^2
x 1 + 25x^2
(c) y = ln(tan(2cos(x
(^2) ) y = ln(tan(2cos(x ))
(^2) ) y = ln(tan(2cos(x ))
(^2) ) )) dy dx
tan(2cos(x^2 ))
· sec^2 (2cos(x
(^2) ) ) · ln 2(2cos(x
(^2) ) ) · (− sin(x^2 )) · 2 x
(d) y =
x + eπ cos 4 + sin^5 (6x)
y =
x + eπ cos 4 + sin^5 (6x)
y =
x + eπ cos 4 + sin^5 (6x)
Note that eπ^ and cos 4 are constants.
dy dx
(1)(cos 4 + sin^5 (6x)) − (x + eπ^ )(5 sin^4 (6x) · cos(6x) · 6) (cos 4 + sin^5 (6x))^2
Recall that sin^5 (6x) = (sin(6x))^5.
- Consider the curve defined by x^3 + y^3 =
x^3 + y^3 = xy
x^3 + y^3 = xy
xy (known as the Folium of Descartes).
(a) Find dy/dxdy/dxdy/dx. Use implicit differentiation.
3 x^2 + 3y^2
dy dx
y +
x
dy dx 3 y^2
dy dx
x
dy dx
y − 3 x^2
dy dx
3 y^2 −
x
y − 3 x^2
dy dx
9 2 y^ −^3 x
2 3 y^2 − 92 x
(b) Verify that the point (1,2) is on the curve above. We must check to see if the values x = 1 and y = 2 satisfy the equation above.
x^3 + y^3 ? =
xy
13 + 2^3 ? =
? = 9
Thus, the point (1,2) is on the curve.
(c) Find the equation of the tangent line at the point (1,2). We want y = mx + b.
m =
9 2 ·^2 −^3 ·^1 2 3 · 22 − 92 · 1
, so y =
x + b.
Now plug in x = 1 and y = 2 to find b. 2 =
· 1 + b ⇒
= b
Therefore, we have y =
x +
- Find the following.
(a) an antiderivative of y =
1 − 9 x^2
y = (^) + x^3 + cos(2x) + e^3
1 − 9 x^2
y = +^ x^3 + cos(2x) +^ e^3
1 − 9 x^2
5 arcsin 3x 3
x^4 4
sin 2x 2
(b) tan(arccostan(arccostan(arccos xxx))) (rewritten as an algebraic expression - no trigonometric functions) Let θ = arccos x. That is, θ is the angle whose cosine is x.
x
y
θ
x^2 + y^2 = 1^2 ⇒ y =
1 − x^2
tan(arccos x) = tan θ =
opposite adjacent
y x
1 − x^2 x
- Consider the function f f f(((xxx) =) =) = xxx^444 eeexxx^ with domain all real numbers.
(a) Find the xxx-value(s) of all roots (xxx-intercepts) of fff. The equation x^4 ex^ = 0 means x^4 = 0 (that is, x = 0) or ex^ = 0 (no solution), so the only root is at x = 0. (b) Find the xxx- and yyy-value(s) of all critical points and identify each as a local max, local min, or neither.
f′^ (x) = 4x^3 ex^ + x^4 ex 0 = x^3 ex(4 + x) ⇒ x = 0, − 4 Note that ex^ is never 0.
x < − 4 − 4 < x < 0 4 < x f′^ positive negative positive f ↗ ↘ ↗ y-values: f(−4) = 256e−^4 ≈ 4 .689, f(0) = 0 So, f has a local maximum at (− 4 , 256 e−^4 ) and a local minimum at (0, 0).
(c) Find the xxx- and yyy-value(s) of all global extrema and identify each as a global max or global min. There is a global minimum at (0, 0). There is no global maximum because as x → ∞, f(x) → ∞. Note that as x → −∞, f(x) → 0. You can verify this by using L’Hopital’s Rule on x^4 /e−x. (d) Find the xxx-value(s) of all inflection points.
f′′(x) = 12x^2 ex^ + 4x^3 ex^ + 4x^3 ex^ + x^4 ex^ Use Product Rule on each product in f′(x) above. 0 = ex^ (x^4 + 8x^3 + 12x^2 ) 0 = ex^ x^2 (x^2 + 8x + 12) 0 = ex^ x^2 (x + 2)(x + 6) ⇒ x = 0, − 2 , − 6
x < − 6 − 6 < x < − 2 − 2 < x < 0 0 < x f′′^ positive negative positive positive f concave up concave down concave up concave up
Goal : Maximize volume
Objective function: volume = V = x · x · y = x^2 y
We need to get this down to a function of just one variable, so we use the constraint equation:
total cost =(cost of base) + (cost of two square ends) + (cost of two other sides) 288 = 3 xy + 12 · 2 x^2 + 12 · 2 xy 288 = 27xy + 24x^2 288 − 24 x^2 = 27xy 288 − 24 x^2 27 x = y
Substituting this back into the objective function gives
V = x^2 y = x^2 ·
288 − 24 x^2 27 x
= x ·
288 − 24 x^2 27
(288x − 24 x^3 ).
Now that we have V as a function of just one variable, we find its maximum.
V ′(x) =
(288 − 72 x^2 )
0 =
(288 − 72 x^2 )
0 = (288 − 72 x^2 ) 72 x^2 = 288
x^2 =
x = 2 We discard x = −2 because lengths must be nonnegative.
Since V ′^ is positive for x < 2 and negative for 2 < x, we know that the maximum occurs at x = 2.
And y =
288 − 24 x^2 27 x
, so the dimensions are 2 by 2 by
