18.01 Solutions, additional problems

• §2.5 #22

Consider the limit

lim_{θ→0} \frac{1 − cos θ}{θ^{2}} .

a) Use a calculator to construct a table of values of the function for small θ’s, and thereby form a conjecture about the value of this limit. Solution: A table of approximate values is:

\begin{matrix}θ & (1 − cos θ)/θ^{2} \\ 0.3 & 0.4963 \\ 0.2 & 0.4983 \\ 0.1 & 0.4996 \\ 0.01 & 0.499996. \end{matrix}

So we conjecture the limit is 1/2.

b) Prove your conjecture. Solution:

lim_{θ→0} \frac{1 − cos θ}{θ^{2}} = lim_{θ→0} \frac{(1 + cos θ)(1 − cos θ)}{(1 + cos θ)θ^{2}}
= lim_{θ→0} \frac{1 − cos^{2} θ}{(1 + cos θ)θ^{2}}
= lim_{θ→0} \frac{sin^{2} θ}{(1 + cos θ)θ^{2}}
= lim_{θ→0} \frac{1}{1 + cos θ} · (lim_{θ→0} \frac{sin θ}{θ} )^{2}
= \frac{1}{2}

since lim_{θ→0} \frac{sin θ}{θ} = 1.

• §3.5 #3 Find dy/dx by implicit differentiation, where x = y − y^{7}.

Solution: Differntiating both sides of the equation with respect to x gives

1 = \frac{dy}{dx} − 7y^{6} \frac{dy}{dx} = (1 − 7y^{6})\frac{dy}{dx},

so dy/dx = 1/(1 − 7y^{6}).

• §3.5 #10 Find dy/dx by implicit differentiation, where \sqrt{xy} + 4 = y.

Solution:

\frac{dy}{dx} = \frac{d}{dx} ((xy)^{1/2} + 4)
= \frac{1}{2}(xy)^{−1/2} · \frac{d}{dx}(xy)
= \frac{1}{2}(xy)^{−1/2}(y + x\frac{dy}{dx})
= \frac{1}{2}(xy)^{−1/2}y + \frac{1}{2}(xy)^{−1/2}x\frac{dy}{dx}
= \frac{1}{2}x^{−1/2}y^{1/2} + \frac{1}{2}x^{1/2}y^{−1/2} \frac{dy}{dx}.

Solving for dy/dx, we get

(1 − \frac{1}{2}x^{1/2}y^{−1/2})\frac{dy}{dx} = \frac{1}{2}x^{−1/2}y^{1/2};

\frac{dy}{dx} = \frac{\frac{1}{2}x^{−1/2}y^{1/2}}{1 − \frac{1}{2}x^{1/2}y^{−1/2}}
= \frac{x^{−1/2}y^{1/2}}{2 − x^{1/2}y^{−1/2}}
= \frac{y}{2\sqrt{xy} − x}

(multiplying the numerator and denominator by \sqrt{xy}).

• §3.5 #28 Find the derivative of

\sqrt{\frac{x^{2} + 3}{x^{2} − 3}}.

Solution:

\frac{d}{dx} (\frac{x^{2} + 3}{x^{2} − 3})^{1/2} = \frac{1}{2} (\frac{x^{2} + 3}{x^{2} − 3})^{1/2} \frac{d}{dx} (\frac{x^{2} + 3}{x^{2} − 3})
= \frac{1}{2} (\frac{x^{2} + 3}{x^{2} − 3})^{1/2} (\frac{2x(x^{2} − 3) − 2x(x^{2} + 3)}{(x^{2} − 3)^{2}})
= \frac{1}{2} (\frac{x^{2} + 3}{x^{2} − 3})^{1/2} (\frac{−12x}{(x^{2} − 3)^{2}} )
= \frac{−6x(x^{2} + 3)^{1/2}}{(x^{2} − 3)^{5/2}} .

• §3.5 #29b Find the equation of the tangent to x^{4} + 16y^{4} = 32 at (2, 1).

Solution: Implicit differentiation gives

4x^{3} + 64y^{3}(dy/dx) = 0,

so dy/dx = −x^{3}/16y^{3}. The slope of the tangent line at (2, 1) is therefore −2^{3}/16 · 1^{3} = −1/2. The equation of the tangent line is therefore y − 1 =

−\frac{1}{2}(x − 2), or

y = −\frac{x}{2} + 2.

• §3.5 #39

Find dy/dx by implicit differentiation if sin y = xy.

Solution:

cos y \frac{dy}{dx} = y + x\frac{dy}{dx};
(cos y − x)\frac{dy}{dx} = y;
\frac{dy}{dx} = \frac{y}{cos y − x}.

• §3.6 #2a

Calculate y′′ if y = \frac{x}{1 − x}.

Solution:

y′ = \frac{(1 − x) − x(−1)}{(1 − x)^{2}} = (1 − x)^{−2};

y′′ = 2(1 − x)^{−3}.

• §3.6 #2e

Calculate \frac{d^{500}}{dx^{500}} (x^{131} − 3x^{79} + 4).

Solution: Since the function is a polynomial of degree 131, and 500 > 131, we have

\frac{d^{500}}{dx^{500}} (x^{131} − 3x^{79} + 4) = 0.

• §3.5 #6

Find the values of y′, y′′ and y′′′ at the point (4, 3) on the circle x^{2} + y^{2} = 25.

Solution: Implicit differentiation gives 2x + 2y(dy/dx) = 0, so y′ = −x/y. Differentiating this gives

y′′ = \frac{d}{dx}(−xy^{−1}) = −y^{−1} + xy^{−2}y′.

Substituting y′ = xy^{−1} gives y′′ = −y^{−1} + xy^{−2}(−xy^{−1}), and simplifying this using the equation x^{2} + y^{2} = 25 gives

y′′ = −y^{−1} − x^{2}y^{−3} = −y^{−3}(x^{2} + y^{2}) = −25y^{−3}.

Differentiating again gives

y′′′ = 75y^{−4}y′ = −75xy^{−5}.

At the point (4, 3), the values are

y′ = −4/3; y′′ = −25/27; y′′′ = −100/81.

• §3.5 #7a

If s = 8t^{2} − \frac{1}{t} (t > 0) is the position of a moving body at time t, find the time, position and velocity at each moment when the acceleration is zero.

Solution: The velocity is

v = s′ = 16t + t^{−2},

and the acceleration is

a = s′′ = v′ = 16 − 2t^{−3}.

So the acceleration is zero when 16 = 2t^{−3}, so t^{3} = 1/8, so t = 1/2. The position at time t = 1/2 is s = 0, and the velocity is v = 12.