3.9E: Exercises for Section 3.8

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In exercises 1 - 10, use implicit differentiation to find $\dfrac{dy}{dx}$.

1) $x^2−y^2=4$

2) $6x^2+3y^2=12$

Answer: $\dfrac{dy}{dx}=\dfrac{−2x}{y}$

3) $x^2y=y−7$

4) $3x^3+9xy^2=5x^3$

Answer: $\dfrac{dy}{dx}=\dfrac{x}{3y}−\dfrac{y}{2x}$

5) $xy−\cos(xy)=1$

6) $y\sqrt{x+4}=xy+8$

Answer: $\dfrac{dy}{dx}=\dfrac{y−\dfrac{y}{2\sqrt{x+4}}}{\sqrt{x+4}−x}$

7) $−xy−2=\frac{x}{7}$

8) $y\sin(xy)=y^2+2$

Answer: $\dfrac{dy}{dx}=\dfrac{y^2\cos(xy)}{2y−\sin(xy)−xy\cos(xy)}$

9) $(xy)^2+3x=y^2$

10) $x^3y+xy^3=−8$

Answer: $\dfrac{dy}{dx}=\dfrac{−3x^2y−y^3}{x^3+3xy^2}$

For exercises 11 - 16, find the equation of the tangent line to the graph of the given equation at the indicated point. Use a calculator or computer software to graph the function and the tangent line.

11) [T] $x^4y−xy^3=−2, \quad (−1,−1)$

12) [T] $x^2y^2+5xy=14,\quad (2,1)$

Answer

$y=−\frac{1}{2}x+2$

$The graph has a crescent in each of the four quadrants. There is a straight line marked T(x) with slope −1/2 and y intercept 2.$

13) [T] $\tan(xy)=y,\quad \left(\frac{π}{4},1\right)$

14) [T] $xy^2+\sin(πy)−2x^2=10, \quad (2,−3)$

Answer

$y=\frac{1}{π+12}x−\frac{3π+38}{π+12}$

$The graph has two curves, one in the first quadrant and one in the fourth quadrant. They are symmetric about the x axis. The curve in the first quadrant goes from (0.3, 5) to (1.5, 3.5) to (5, 4). There is a straight line marked T(x) with slope 1/(π + 12) and y intercept −(3π + 38)/(π + 12).$

15) [T] $\dfrac{x}{y}+5x−7=−\frac{3}{4}y, \quad (1,2)$

16) [T] $xy+\sin(x)=1,\quad \left(\frac{π}{2},0\right)$

Answer

$y=0$

$The graph starts in the third quadrant near (−5, 0), remains near 0 until x = −4, at which point it decreases until it reaches near (0, −5). There is an asymptote at x = 0. The graph begins again near (0, 5) decreases to (1, 0) and then increases a little bit before decreasing to be near (5, 0). There is a straight line marked T(x) that coincides with y = 0.$

17) [T] The graph of a folium of Descartes with equation $2x^3+2y^3−9xy=0$ is given in the following graph.

$A folium is graphed which has equation 2x3 + 2y3 – 9xy = 0. It crosses over itself at (0, 0).$

a. Find the equation of the tangent line at the point $(2,1)$. Graph the tangent line along with the folium.

b. Find the equation of the normal line to the tangent line in a. at the point $(2,1)$.

18) For the equation $x^2+2xy−3y^2=0,$

a. Find the equation of the normal to the tangent line at the point $(1,1)$.

b. At what other point does the normal line in a. intersect the graph of the equation?

Answer: a. $y=−x+2$
b. $(3,−1)$

19) Find all points on the graph of $y^3−27y=x^2−90$ at which the tangent line is vertical.

20) For the equation $x^2+xy+y^2=7$,

a. Find the $x$-intercept(s).

b.Find the slope of the tangent line(s) at the $x$-intercept(s).

c. What does the value(s) in part b. indicate about the tangent line(s)?

Answer: a. $\left(±\sqrt{7},0\right)$
b. $−2$
c. They are parallel since the slope is the same at both intercepts.

21) Find the equation of the tangent line to the graph of the equation $\sin^{−1}x+\sin^{−1}y=\frac{π}{6}$ at the point $\left(0,\frac{1}{2}\right)$.

22) Find the equation of the tangent line to the graph of the equation $\tan^{−1}(x+y)=x^2+\frac{π}{4}$ at the point $(0,1)$.

Answer: $y=−x+1$

23) Find $y′$ and $y''$ for $x^2+6xy−2y^2=3$.

24) [T] The number of cell phones produced when $x$ dollars is spent on labor and $y$ dollars is spent on capital invested by a manufacturer can be modeled by the equation $60x^{3/4}y^{1/4}=3240$.

a. Find $\frac{dy}{dx}$ and evaluate at the point $(81,16)$.

b. Interpret the result of a.

Answer: a. $\frac{dy}{dx}=−0.5926$
b. When $81 is spent on labor and $16 is spent on capital, the amount spent on capital is decreasing by $0.5926 per $1 spent on labor.

25) [T] The number of cars produced when $x$ dollars is spent on labor and $y$ dollars is spent on capital invested by a manufacturer can be modeled by the equation $30x^{1/3}y^{2/3}=360$.

(Both $x$and $y$ are measured in thousands of dollars.)

a. Find $\frac{dy}{dx}$ and evaluate at the point $(27,8)$.

b. Interpret the result of part a.

26) The volume of a right circular cone of radius $x$ and height $y$ is given by $V=\frac{1}{3}πx^2y$. Suppose that the volume of the cone is $85π\,\text{cm}^3$. Find $\dfrac{dy}{dx}$ when $x=4$ and $y=16$.

Answer: $\dfrac{dy}{dx} = −8$

For exercises 27 - 28, consider a closed rectangular box with a square base with side $x$ and height $y$.

27) Find an equation for the surface area of the rectangular box, $S(x,y)$.

28) If the surface area of the rectangular box is 78 square feet, find $\dfrac{dy}{dx}$ when $x=3$ feet and $y=5$ feet.

Answer: $\dfrac{dy}{dx} = −2.67$

In exercises 29 - 31, use implicit differentiation to determine $y′$. Does the answer agree with the formulas we have previously determined?

29) $x=\sin y$

30) $x=\cos y$

Answer: $y′=−\dfrac{1}{\sqrt{1−x^2}}$

31) $x=\tan y$