Solved Problems http://solvedproblems.wordpress.com Worked out exercises in algebra, geometry, trigonometry, calculus, etc. Sun, 03 Aug 2008 13:45:09 +0000 http://wordpress.com/ en hourly 1 http://www.gravatar.com/blavatar/ce7f4ad763eeb53a8c353e84afb8c8be?s=96&d=http://s.wordpress.com/i/buttonw-com.png Solved Problems http://solvedproblems.wordpress.com Some more algebra problems http://solvedproblems.wordpress.com/2008/08/03/some-more-algebra-problems/ http://solvedproblems.wordpress.com/2008/08/03/some-more-algebra-problems/#comments Sun, 03 Aug 2008 13:45:09 +0000 ldvallejo http://solvedproblems.wordpress.com/?p=61 ]]>

Sorry, I will not be able to answer all of my semi-promised exercises. The UPCAT sked is so toxic, I am writing this midnight and I have to be at UP at 530 am! suicide, huh. haha

Anyway, I’ll just work on some examples of my own and just read between the lines why.

An inlet pipe can fill the pool in 3 hours while a drain pipe will drain the pool in 8 hours. If the inlet pipe is turned on and the outlet pipe is accidentally also turned on after one hour that the inlet pipe gushing, how many more minutes would it take for the pool to be filled halfway?

Soln: rate of inlet pipe is 1/3

rate of outlet pipe is -1/8

Since the inlet pipe has been gushing for 1 hour, it would have done 1 times 1/3 or 1/3 of the work. The equation would look like this:

1 (1/3) + (1/3 + -1/ 8 ) t = 1/2

where t is the time wherein the pool becomes halfway full. You know of course how to solve for this.

Find the value of s such that the perimeter of the the equilateral triangle with side s is equal to the area of the same equilateral triangle with side equal to s.

Soln: We know that the perimeter of an equilateral triangle with side equal to s is 3s and you may verify that the area of the triangle with side equal to s is sqrt(3) s^2 divided by 4. Equate the two and you would get the correct answer.

]]> http://solvedproblems.wordpress.com/2008/08/03/some-more-algebra-problems/feed/ 6 ldvallejo Equations of Lines in 3D http://solvedproblems.wordpress.com/2008/04/30/equations-of-lines-in-3d/ http://solvedproblems.wordpress.com/2008/04/30/equations-of-lines-in-3d/#comments Wed, 30 Apr 2008 10:38:27 +0000 ldvallejo http://solvedproblems.wordpress.com/?p=60 ]]>

Let l be the line represented by the symmetric equations

\displaystyle\frac{x-2}{4} = \frac{y+3}{-2}=\frac{z-1}{7}

1. Find the point of intersection of l and the plane 5x-y+2z=12.

2. Find the distance between P(0,1,0) and l.

3. The acute angle between l and the line m: \hspace{.3in} x=1+t; \hspace{.2in} y=3-2t; \hspace{.2in} z=4t.

Solution:

1. Note that transforming l in its parametric form of equations gives us:

x=4t+2; \hspace{.2in} y=2t-3; \hspace{.2in} z=7t+1

Plugging the values x, y and z into the equation of the plane gives us an equation of one variable t:

5(4t+2)-(-2t-3)+2(7t+1)=12

Solving the above equation gives us the value t=-1/12.

Finally, plug this value of parameter to the equation of the line.

x=4(-1/12)+2 = 5/3

y=-2(-1/12)-3 = -17/6

z=7(-1/12)+1 = 5/12

Thus, the point of intersection is the point (5/3,-17/6,5/12).

2. A parallel vector to l is <4,-2,7> =: R and a point on l is Q(2,-3,1). (It is hard to put a drawing here so, just do the visualization yourself). The plan is to construct a right triangle with hypotenuse parallel to the vector PQ and one leg parallel to the R . We will find now the scalar projection of PQ onto R and the magnitude of PQ.

sp(PQ,R) = \displaystyle\frac{<2,-4,1> \cdot <4,-2,7>}{ \sqrt{16+4+49}} = \frac{23}{\sqrt{69}}

\left\| PQ \right\| = \sqrt{4+16+1}= \sqrt{21}

This is still the length of the leg parallel to R and the length of the hypotenuse. FInally, we use Pythagorean Theorem to find our desired distance.

dist(P,l) = \sqrt{21 - \frac{23^2}{69}}=\sqrt{\frac{920}{69}}

3. The parallel vector to l is <4,-2,7> =: R while the parallel vector to m is <1,-2,4> =: S . The acute angle is then found from the formula of the dot product

\theta = \arccos\displaystyle\frac{R\cdot S}{\left\|R\right\| \cdot \left\| S \right\|}

Meanwhile,

R\cdot S = <4,-2,7> \cdot <1,-2,4> = 36

\left\|R\right\| = \sqrt{16+4+49}=\sqrt{69}

\left\| S \right\| = \sqrt{1+4+16} = \sqrt{21}

Plugging all of these values gives us

\theta = \arccos\displaystyle\frac {36}{\sqrt{69} \cdot \sqrt{21}}

]]> http://solvedproblems.wordpress.com/2008/04/30/equations-of-lines-in-3d/feed/ 1 ldvallejo Linear Independence http://solvedproblems.wordpress.com/2008/04/06/linear-independence/ http://solvedproblems.wordpress.com/2008/04/06/linear-independence/#comments Sun, 06 Apr 2008 07:40:13 +0000 ldvallejo http://solvedproblems.wordpress.com/?p=59 ]]>

Let V be a finite vector space over the set of rationals, and F be a linear transformation from V to V. Suppose x, y and z are vectors in V such that

(i) F(x) = y

(ii) F(y) = z

(iii) F(z) = x+y

Suppose x is nonzero. Show that x, y and z are linearly independent vectors.

Proof: We first want to show that x and y are linearly independent. Suppose not. Then there exists a nonzero rational a such that y = ax.

Applying F to this equation yields z = ay = a(ax) = a^2 x. Thus the three vectors can be written x,ax and a^2x .

Now, consider F(z) = x + y = x + ax = x(1+a). On the other hand, F(z) = F(a^2 x) = a^2 F(x) = a^2y = a^3x.

So that by transitivity, (a^3-a-1) x=0. Since x is nonzero, a^3-a-1=0. However, you can check that this polynomial does not have rational roots (e.g. by remainder theorem). Thus a does not exist and therefore x and y are linearly independent.

It is now left to show that z is linearly independent to x and y. Suppose there exists rational numbers a and b such that z = ax+by. Applying F twice gives us y+z = az + b(x+y). That is, (1-a)z = bx + (b-1)y.

a cannot be equal to 1, otherwise, x and y would be linearly dependent, a contradiction. Thus we can write z = \displaystyle\frac{b}{1-a}x + \frac{b-1}{1-a}y. But x and y are linearly independent so a representation of z in terms of them will be unique. Finally, we will now compare coefficients and come up with a = \displaystyle\frac{b}{1-a} and b=\displaystyle\frac{b-1}{1-a}, which simplify to a-a^2 = b and ab =1. Solving for a and b, we get a^3-a^2+1=0 and b^3-b+1. These polynomials have no rational roots. Thus, by the same argument as of that of the above claim, a and b do not exist and therefore we can conclude that x,y and z are linearly independent.

]]> http://solvedproblems.wordpress.com/2008/04/06/linear-independence/feed/ 0 ldvallejo Area, Volume and Perimeter http://solvedproblems.wordpress.com/2008/03/23/area-volume-and-perimeter/ http://solvedproblems.wordpress.com/2008/03/23/area-volume-and-perimeter/#comments Sun, 23 Mar 2008 08:48:18 +0000 wMw http://solvedproblems.wordpress.com/?p=57 ]]>

Consider the regions \displaystyle \mathcal{R}_{1}\,\mathrm{and}\,\mathcal{R}_{2} enclosed by the curves \displaystyle y=x^{3}+\frac{1}{2} and \displaystyle x=-\left(y-\frac{1}{2}\right)^{2}. Set-up the integrals representing the following:

  1. The perimeter of \displaystyle \mathcal{R}_{2}.
  2. The total area of the regions \displaystyle \mathcal{R}_{1} and \displaystyle \mathcal{R}_{2}.
  3. The volume of the solid generated when \displaystyle \mathcal{R}_{1} is rotated about the line \displaystyle y=-\frac{1}{2} using the method of washers.
  4. The volume of the solid generated when \displaystyle \mathcal{R}_{2} is rotated about the line x=-1 using the method of cylindrical shells.

Solution

The regions \displaystyle \mathcal{R}_{1}\,\mathrm{and}\,\mathcal{R}_{2} are shown below.

The perimeter of \displaystyle \mathcal{R}_{2} consists of the horizontal segment of length 2, and the lengths of partial arcs of the two curves. We use the arc length formula to compute the perimeter \mathcal{P}. Since \displaystyle y'=3x^{2} and \displaystyle x'=-2\left(y-\dfrac{1}{2}\right)^{2}, we have

\displaystyle \mathcal{P} = 2+\int_{0}^{1}\sqrt{1+9x^{4}}\,dx+\int_{\frac{1}{2}}^{\frac{3}{2}}\sqrt{1+4\left(y-\dfrac{1}{2}\right)^{2}}\,dy

Now, to compute the total area of the two regions, we consider horizontal strips:

In the region \displaystyle \mathcal{R}_{1}, the function at the right is \displaystyle x=-\left(y-\dfrac{1}{2}\right)^2 and the function at the left is \displaystyle x=\sqrt[3]{y-\dfrac{1}{2}}.

In the region \displaystyle \mathcal{R}_{2}, the function at the right is \displaystyle x=\sqrt[3]{y-\dfrac{1}{2}} and the function at the left is \displaystyle x=-\left(y-\dfrac{1}{2}\right)^2.

Therefore, the total area is given by

\displaystyle \mathcal{A} = \int_{-\frac{1}{2}}^{\frac{1}{2}}\left[-\left(y-\dfrac{1}{2}\right)^2-\sqrt[3]{y-\dfrac{1}{2}}\right]\,dy

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\mathcal{A}} + \int_{\frac{1}{2}}^{\frac{3}{2}}\left[\sqrt[3]{y-\dfrac{1}{2}}+\left(y-\dfrac{1}{2}\right)^2\right]\,dy

We now turn to the volumes of the two solids generated. Note that for number 3, we need representations for the outer and inner radii of the washers for the generated solid. But first, observe that we can express \displaystyle x=-\left(y-\dfrac{1}{2}\right)^2 as \displaystyle y=\dfrac{1}{2}-\sqrt{-x} in the region \displaystyle \mathcal{R}_{1}. Thus, the volume of the solid generated by revolving \displaystyle \mathcal{R}_{1} about y=-\dfrac{1}{2} is given by

\displaystyle \mathcal{V}_{\mathcal{R}_{1}}=\pi\int_{-1}^{0}\left[ \left(x^{3}+\dfrac{1}{2}+\dfrac{1}{2}\right)^{2}-\left(\dfrac{1}{2}-\sqrt{-x}+\dfrac{1}{2}\right)^{2}\right]\,dx

or

\displaystyle \mathcal{V}_{\mathcal{R}_{1}}=\pi\int_{-1}^{0}\left[\left( x^{3}+1\right)^{2}-\left( 1-\sqrt{-x}\right)^{2}\right]\,dx.

Next, to obtain the volume of the solid generated by rotating \displaystyle \mathcal{R}_{2} about x=-1, we note that the upper branch of \displaystyle x=-\left(y-\dfrac{1}{2}\right)^2 can be expressed as \displaystyle y=\dfrac{1}{2}+\sqrt{-x} in the region \displaystyle \mathcal{R}_{2}. Moreover, notice that the upper function varies in the region. Hence, by cylindrical shells, we express the volume as

\displaystyle \mathcal{V}_{\mathcal{R}_{2}}=2\pi\int_{-1}^{0}\left(x+1\right)\left[\frac{3}{2}-\left(\frac{1}{2}+\sqrt{-x}\right)\right]\,dx

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\mathcal{V}_{\mathcal{R}_{2}}}+2\pi\int_{0}^{1}(x+1)\left[\frac{3}{2}-\left(x^{3}+\frac{1}{2}\right)\right]\,dx

or

\displaystyle \mathcal{V}_{\mathcal{R}_{2}}=2\pi\left[\int_{-1}^{0}\left(x+1\right)\left(1-\sqrt{-x}\right)\,dx+\int_{0}^{1}(x+1)\left(1-x^{3}\right)\,dx\right].

]]> http://solvedproblems.wordpress.com/2008/03/23/area-volume-and-perimeter/feed/ 5 wMw Related Rates http://solvedproblems.wordpress.com/2008/03/23/related-rates/ http://solvedproblems.wordpress.com/2008/03/23/related-rates/#comments Sun, 23 Mar 2008 07:19:53 +0000 wMw http://solvedproblems.wordpress.com/?p=53 ]]>

A particle moves along the curve y=\ln x so that its abscissa is increasing at a rate of 2 units per second. At what rate is the particle moving away from the origin as it passes through the point (e,1)?

Solution

Let R be the distance between the particle and the origin, and t denote time in seconds. Using the distance formula, we can express R as a function of x:

\displaystyle R = \sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+\left(\ln x\right)^{2}}

We are to find \displaystyle \frac{dR}{dt} at (e,1) given that \displaystyle \frac{dx}{dt}=2. Differentiating the above equation implicitly with respect to t, we obtain

\displaystyle \frac{dR}{dt}=\frac{\dfrac{2\ln x}{x}\dfrac{dx}{dt}+2x\dfrac{dx}{dt}}{2\sqrt{x^{2}+\left(\ln x\right)^{2}}}.

Hence,

\displaystyle \left.\frac{dR}{dt}\right|_{(e,1)}=\frac{\dfrac{4}{e}+4e}{2\sqrt{1+e^{2}}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\left.\frac{dR}{dt}\right|_{(e,1)}}=\frac{2\left(\dfrac{e^{2}+1}{e}\right)}{\sqrt{1+e^{2}}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\left.\frac{dR}{dt}\right|_{(e,1)}}=\frac{2\sqrt{1+e^{2}}}{e}

]]> http://solvedproblems.wordpress.com/2008/03/23/related-rates/feed/ 0 wMw Largest Rectangle in Some Region in the First Quadrant http://solvedproblems.wordpress.com/2008/03/23/largest-rectangle-in-some-region-in-the-first-quadrant/ http://solvedproblems.wordpress.com/2008/03/23/largest-rectangle-in-some-region-in-the-first-quadrant/#comments Sun, 23 Mar 2008 06:23:15 +0000 wMw http://solvedproblems.wordpress.com/?p=54 ]]>

Find the area of the largest rectangle that can be inscribed in the region bounded by the curve y=e^{-x}, the line x=0 and the positive x-axis.

Solution

Denote by x the point as shown in the figure. Then the rectangle inscribed in the region has dimensions x by e^{-x}. Thus, we need to maximize the area function

\displaystyle A(x)=xe^{-x}

To this end, we compute the function’s derivative:

A'(x)=-xe^{-x}+e^{-x}=e^{-x}\left(1-x\right)

We see that x=1 is the only critical point of the function.

We shall now use the first derivative test to show that the point where x=1 yields a maximum value. Note that e^{-x} is always positive hence the sign of \displaystyle A' depends only on the sign of the factor 1-x. It is now apparent that A' is positive if x<1 and negative if x>1. By the first derivative test, the point x=1 gives a relative maximum value.

The result also follows immediately from the second derivative test. Indeed, A''(x)=-e^{-x}\left(2-x\right) and clearly, A''(1)<0.

We therefore conclude that the largest rectangle that can be inscribed in the region has dimensions 1 and \displaystyle \frac{1}{e} and its area is \displaystyle \frac{1}{e} square units.

]]> http://solvedproblems.wordpress.com/2008/03/23/largest-rectangle-in-some-region-in-the-first-quadrant/feed/ 0 wMw Normal Line Problem http://solvedproblems.wordpress.com/2008/03/23/normal-line-problem/ http://solvedproblems.wordpress.com/2008/03/23/normal-line-problem/#comments Sun, 23 Mar 2008 05:01:04 +0000 wMw http://solvedproblems.wordpress.com/?p=52 ]]>

Find the equation of the line normal to the curve \displaystyle y=\tanh \left[\sin^{-1} \left( x-\sqrt{2}\right) \right] at the point where \displaystyle x=\sqrt{2}.

Solution

We first compute the derivative of the function using the Chain Rule:

\displaystyle \frac{dy}{dx} = \mathrm{sech}^{2}\left[ \sin^{-1}\left( x-\sqrt{2} \right) \right]\cdot \frac{1}{\sqrt{1-\left(x-\sqrt{2}\right)^{2}}}

At the point on the curve where x = \sqrt{2}, y=\tanh \left(\sin^{-1} 0\right)=0 and the slope of the tangent line is

\displaystyle \left.\frac{dy}{dx}\right|_{x=\sqrt{2}} = \mathrm{sech}^{2} 0 \cdot 1 = 1

Thus, the slope of the line normal to the curve at the indicated point is -1. The desired equation is

\displaystyle y=-1\left( x - \sqrt{2}\right)

or

\displaystyle y=\sqrt{2}-x.

]]> http://solvedproblems.wordpress.com/2008/03/23/normal-line-problem/feed/ 0 wMw Integrals Involving Transcendental Functions http://solvedproblems.wordpress.com/2008/03/23/integrals-involving-transcendental-functions/ http://solvedproblems.wordpress.com/2008/03/23/integrals-involving-transcendental-functions/#comments Sat, 22 Mar 2008 18:37:52 +0000 wMw http://solvedproblems.wordpress.com/?p=51 ]]>

Evaluate the following integrals.

  1. \displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx
  2. \displaystyle \int\frac{2^{x}\: dx}{\left(2^{x}+1\right)\sqrt{4^{x}+2^{x+1}-8}}
  3. \displaystyle \int\frac{dx}{x+4\sqrt{x}+13}

Solution

We use the substitution rule in the following computations.

For the first integral, let u =\ln\left(\sin x\right). Then du=\dfrac{1}{\sin x}\cdot\cos x\: dx=\cot x\: dx. Moreover, if x=\dfrac{\pi}{6}, u=\ln\left(\dfrac{1}{2}\right)=-\ln2 and if x=\dfrac{\pi}{4}, u=\ln\left(\dfrac{\sqrt{2}}{2}\right)=-\dfrac{1}{2}\ln2. Thus,

\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\ln^{2}\left(\sin x\right)\cot x\: dx

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx } = \int_{-\ln2}^{-\frac{1}{2}\ln2}u^{2}\: du

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx } = \left.\frac{u^{3}}{3}\right|_{-\ln2}^{-\frac{1}{2}\ln2}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx } = \frac{1}{3}\left(-\frac{1}{8}\ln^{3}2+\ln^{3}2\right)

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\ln^{2}\left(\sin x\right)}{\tan x}\: dx } = \frac{7}{24}\ln^{3}2

In the second integral, note that

\displaystyle 4^{x}+2^{x+1}-8=2^{2x}+2\cdot2^{x}+1-9=\left(2^{x}+1\right)^{2}-3^{2}.

So we suspect that the given integral yields the inverse secant function. Indeed, if we make the substitution u=2^{x}+1, then du=2^{x}\ln2dx and thus

\displaystyle \int\frac{2^{x}\: dx}{\left(2^{x}+1\right)\sqrt{4^{x}+2^{x+1}-8}} = \frac{1}{\ln2}\int\frac{du}{u\sqrt{u^{2}-3^{2}}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{2^{x}\:dx}{\left( 2^{x}+1\right) \sqrt{4^{x}+2^{x+1}-8}}} = \frac{1}{\ln2}\cdot\frac{1}{3} \sec^{-1} \left( \frac{u}{3}\right) +C

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{2^{x}\:dx}{\left( 2^{x}+1\right) \sqrt{4^{x}+2^{x+1}-8}}} = \frac{1}{3\ln2}\sec^{-1} \left( \frac{2^{x}+1}{3}\right) +C.

For the last integral, let’s make the substitution u=\sqrt{x}. Then du=\dfrac{1}{2\sqrt{x}}\, dx or dx=2udu. So we have

\displaystyle \int\frac{dx}{x+4\sqrt{x}+13} = \int\frac{2u\, du}{u^{2}+4u+13}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{dx}{x+4\sqrt{x}+13}} = \int\frac{2u+4}{u^{2}+4u+13}\,du-\int\frac{4\,du}{u^{2}+4u+13}.

Observe that the first part above yields the natural logarithmic function while the second part involves the inverse tangent function. In fact, if we let v=u^{2}+4u+13, then dv=\left(2u+4\right)du. Thus,

\displaystyle \int\frac{2u+4}{u^{2}+4u+13}\, du = \int\frac{dv}{v}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{2u+4}{u^{2}+4u+13}\, du} = \ln\left|v\right|+C_{1}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{2u+4}{u^{2}+4u+13}\, du} = \ln\left(u^{2}+4u+13\right)+C_{1}.

Meanwhile,

\displaystyle \int\frac{4\, du}{u^{2}+4u+13} = 4\int\frac{du}{\left(u+2\right)^{2}+3^{2}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\int\frac{4\, du}{u^{2}+4u+13}} = \frac{4}{3}\tan^{-1}\left(\frac{u+2}{3}\right)+C_{2}.

Hence, the third integral yields

\displaystyle \ln\left(x+4\sqrt{x}+13\right)-\frac{4}{3}\tan^{-1}\left(\frac{\sqrt{x}+2}{3}\right)+C.

]]> http://solvedproblems.wordpress.com/2008/03/23/integrals-involving-transcendental-functions/feed/ 4 wMw Indeterminate Forms Involving Transcendental Functions http://solvedproblems.wordpress.com/2008/03/23/indeterminate-forms-involving-transcendental-functions/ http://solvedproblems.wordpress.com/2008/03/23/indeterminate-forms-involving-transcendental-functions/#comments Sat, 22 Mar 2008 17:32:17 +0000 wMw http://solvedproblems.wordpress.com/?p=50 ]]>

Find the limits.

  1. \displaystyle \lim_{x\rightarrow1^{+}}\left(\frac{1}{\ln x}-\frac{1}{\cosh^{-1}x}\right)
  2. \displaystyle \lim_{x\rightarrow0^{+}}\left(e^{x}+\sinh x\right)^{{\displaystyle \cot x}}

Solution

Note that we shall use l’Hospital’s Rule in our computations as necessary.

The first limit has the indeterminate form \infty-\infty. We compute the limit as follows:

\displaystyle \lim_{x\rightarrow 1^{+}}\left(\frac{1}{\ln x}-\frac{1}{\cosh^{-1}x}\right) = \lim_{x\rightarrow 1^{+}}\left(\frac{\cosh^{-1}x-\ln x}{\ln x\cdot\cosh^{-1}x}\right)\: \mathrm{Form}:\frac{0}{0}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow 1^{+}} \left( \frac{1}{\ln x} - \frac{1}{\cosh^{-1}x}\right)} = \lim_{x\rightarrow 1^{+}}\left(\frac{\dfrac{1}{\sqrt{x^{2}-1}} - \dfrac{1}{x}}{\dfrac{\ln x}{\sqrt{x^{2}-1}} + \dfrac{\cosh^{-1}x}{x}}\right)

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow 1^{+}} \left( \frac{1}{\ln x} - \frac{1}{\cosh^{-1}x}\right)} = \lim_{x\rightarrow 1^{+}}\frac{x-\sqrt{x^{2}-1}}{x\ln x+\cosh^{-1}x\cdot\sqrt{x^{2}-1}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow 1^{+}} \left( \frac{1}{\ln x} - \frac{1}{\cosh^{-1}x}\right)} = \frac{1}{0^{+}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow 1^{+}} \left( \frac{1}{\ln x} - \frac{1}{\cosh^{-1}x}\right)} = +\infty

Meanwhile, the second limit has the form 1^{\infty}. Observe that

\displaystyle \lim_{x\rightarrow 0^{+}}\cot x\cdot\ln\left(e^{x}+\sinh x\right) = \lim_{x\rightarrow 0^{+}}\frac{\ln\left(e^{x}+\sinh x\right)}{\tan x}\:\left(\mathrm{Form}:\frac{0}{0}\right)

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x_\rightarrow 0^{+}}\cot x\cdot \ln \left( e^{x} +\sinh x\right)} = \lim_{x\rightarrow 0^{+}} \frac{\left(\dfrac{e^{x}+\cosh x}{e^{x}+\sinh x}\right)}{\sec^{2}x}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x_\rightarrow 0^{+}}\cot x\cdot \ln \left( e^{x} +\sinh x\right)} = 2.

Therefore,

\displaystyle \lim_{x\rightarrow0^{+}}\left(e^{x}+\sinh x\right)^{{\displaystyle \cot x}} = e^{\displaystyle \left[ \lim_{x\rightarrow 0^{+}}\ln\left(e^{x}+\sinh x\right)^{\cot x}\right]}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow0^{+}}\left(e^{x}+\sinh x\right)^{{\displaystyle \cot x}}} = e^{\displaystyle \left[ \lim_{x\rightarrow 0^{+}}\cot x\cdot\ln\left(e^{x}+\sinh x\right)\right]}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\lim_{x\rightarrow0^{+}}\left(e^{x}+\sinh x\right)^{{\displaystyle \cot x}}} = e^{2}

]]> http://solvedproblems.wordpress.com/2008/03/23/indeterminate-forms-involving-transcendental-functions/feed/ 0 wMw Logarithmic and Implicit Differentiation http://solvedproblems.wordpress.com/2008/03/23/logarithmic-and-implicit-differentiation/ http://solvedproblems.wordpress.com/2008/03/23/logarithmic-and-implicit-differentiation/#comments Sat, 22 Mar 2008 16:35:49 +0000 wMw http://solvedproblems.wordpress.com/?p=49 ]]>

Solve for \displaystyle \frac{dy}{dx}.

  1. y=\displaystyle \sqrt[5]{\frac{4^{x}\coth\left(e^{-x}\right)}{\sin\left(x^{2}\ln x\right)}}
  2. \displaystyle e^{y}=\cosh\left(e^{x}+\log_{3}y\right)-2^{{\displaystyle x^{2}}}

Solution

If we proceed directly in number 1, we get

\displaystyle \frac{dy}{dx}=\frac{ y^{-4}}{5}\cdot\frac{\sin\left(x^{2}\ln x\right)\cdot\left[4^{x}e^{-x}\mathrm{csch}^{2}\left(e^{-x}\right)+4^{x}\ln4\coth\left(e^{-x}\right)\right]}{\sin^{2}\left(x^{2}\ln x\right)}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\frac{dy}{dx}=} -\frac{y^{-4}}{5}\cdot\frac{4^{x}\coth\left( e^{-x} \right)\cos\left(x^{2}\ln x\right)\left[ x+2x\ln x\right]}{\sin^{2}\left(x^{2}\ln x\right)}

Meanwhile, if we first simplify the expression by exploiting the properties of logarithms, we obtain

\displaystyle \ln y = \frac{1}{5} \left[ x\ln4 + \ln\coth \left( e^{-x}\right) - \ln \sin \left( x^{2}\ln x \right) \right]

from which the derivative is easier to compute compared to the previous:

\displaystyle \frac{dy}{dx}=\frac{1}{5}y\left[\ln4+\frac{e^{-x}\mathrm{csch}^{2}\left(e^{-x}\right)}{\coth\left(e^{-x}\right)}-\frac{\cos\left(x^{2}\ln x\right)\left[x+2x\ln x\right]}{\sin\left(x^{2}\ln x\right)}\right].

In number 2, we just differentiate the equation implicitly with respect to x:

\displaystyle e^{y} \frac{dy}{dx} = \sinh\left( e^{x} + \log_{3} y \right) \cdot \left(e^{x} + \frac{1}{y\ln 3} \frac{dy}{dx} \right) - 2x \ln 2 \cdot 2^{x^2}

Solving for \displaystyle \frac{dy}{dx}, we get

\displaystyle \frac{dy}{dx}=\frac{e^{x}\sinh\left(e^{x}+\log_{3}y\right)-2x \ln 2 \cdot2^{x^{2}}}{e^{y}-\dfrac{\sinh\left(e^{x}+\log_{3}y\right)}{y\ln3}}

\displaystyle \rule{0.0001em}{0.0001ex} \phantom{\frac{dy}{dx}} =\frac{y\ln3\left[e^{x}\sinh\left(e^{x}+\log_{3}y\right)-2x \ln 2 \cdot2^{x^{2}}\right]}{ye^{y}\ln3-\sinh\left(e^{x}+\log_{3}y\right)}.

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