Assignment 20 Query

course Mth 173

???B??????k??O??X?assignment #020

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s?r??????????????Calculus I

07-25-2007

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07:29:23

Query problem 3.6.12 (3d edition 3.6.44). was 4.6.12 derivative of e^( ln(x) + 1)

What is the derivative of the given function?

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RESPONSE -->

The derivative I found was:

(e^(ln x + 1) ) / x

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07:31:09

Explain what rule or rules you used to obtain your derivative and how you used them.

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RESPONSE -->

I first decomposed the original function into two parts:

f = e^g

g = ln x + 1

Then I found the derivatives of each one:

f' = e^g

g' = 1 / x

I then applied the chain rule:

1 / x * e^(ln x + 1)

Rearranged:

( e^(ln x + 1) ) / x

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07:31:57

07-25-2007 07:31:57

** This is a composite with f(z) = e^z and g(x) = ln(x) + 1.

g'(x) = 1/x, f'(z) = e^z.

So the derivative is

(e^(ln(x)+1)) ' = (f(g(x)) ' =

g'(x) * f'(g(x)) =

1/x e^(ln(x)+1)) =

1/x e^(ln(x)) * e^1 = 1/x * x * e = e. **

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NOTES -------> Ah, I didn't work it out that far. Exponents.

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07:32:27

Why is in easier to calculate the derivative of this function if you simplify the function first?

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RESPONSE -->

Because it is easier to keep track of two small functions than one large messy function.

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07:33:22

** e^(ln x + 1) = e^(ln x) * e^1 = x * e or just e * x.

e is a constant so the derivative of e * x is e * 1 = e. **

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RESPONSE -->

I see.

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07:36:27

Query problem 3.6.43 (3d edition 3.6.44) was 4.6.30 y = ln(x) at x = 1

What is the equation of the tangent line at the given point?

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RESPONSE -->

I found the tangent line equation to be:

y = x - 1

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07:36:34

** tan line at x = 1 passes through the corresponding graph point (1, ln(1) ) = (1, 0).

Slope of tan line equals derivative: y' = 1/x; at x = 1 we have y' = 1.

So line has slope 1 and passes through (1, 0).

The equation of the line is y - 0 = 1 * (x - 1), or y = x-1. **

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RESPONSE -->

Ok.

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07:37:34

What are your approximations to ln(1.1) and ln(2), based on the tangent line?

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RESPONSE -->

My approximations:

ln(1.1): y = 1.1 - 1 = 0.1

ln(2): y = 2 - 1 = 1

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07:37:47

** to get approx ln(1.1): y = 1.1 - 1 = .1.

to get approx ln(1.2): y = 1.2 - 1 = .2.

Actual values are

ln(1.1)=.095 and

ln(1.2)=.18.

Note that both are a little below the approximations given by the tangent line. **

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RESPONSE -->

Ok.

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07:40:48

In terms of the graph, explain whether your approximations are larger or smaller than the true values.

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RESPONSE -->

Both of my approximations are larger than the true values. This is becaue the graph of ln(x) is concave down near x=1 and the tangent line moves upward with a constant slope.

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07:40:55

The value at y=x-1 is higher than the value at y=lnx because the slope of the tangent line works well for values very close to the original point, but there is a limit to its accuracy after moving away so far from the point

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RESPONSE -->

Ok.

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07:41:02

** Since the graph of y = ln(x) is always concave downward, it will always fall below its tangent-line approximation. **

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RESPONSE -->

Ok.

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07:41:11

** the graph of the natural log function is concave downward, so the tan line must live above the graph of the actual function. Thus the approximation given by the tangent line will always be high. **

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RESPONSE -->

Ok.

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07:42:21

Query problem 3.7.16 (was 3.7.9 was 4.7.6) e^(x^2) + ln y = 0

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RESPONSE -->

I found the derivative to be:

dy/dx = 2xy e^(x^2)

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07:43:48

When differentiating with respect to x any x terms are differentiated as usual, to differentiate y with respect to x assumes that y is a function of x, in which case its derivative with respect to x is the unspecified quantity dy/dx.

The derivative of a function of y is therefore the derivative of a composite. For example cos(y) is the composite of f(y) = cos(y) with the function g(x) = y(x), whose form is not specified. Do g ' (x) * f ' (g(x) ) is y ' (x) f ' (y ( x) ); in the present example this would be y ' (x) * ( -sin(y(x)), which we would write -sin(y) y ' with the understanding that y stands for y(x) and y ' for y ' (x).

When we use implicit differentiation to solve for y ' we will typically get some terms containing y ' as a factor and others which don't. We separate these terms algebraically, with the y ' terms on one side and everything else on the other. We then factor out y ' and divide both sides by the other factor to obtain an expression for y ' in terms of y and x.

MORE SPECIFIC SOLUTION:

The derivative of the equation with respect to x is 2x e^(x^2) + 1/y * y ' = 0. Solving this for y ' we get

1/y * y ' = 2 x e^(x^2) so that

y ' = 2x e^(x^2) * y.

To see why the derivative of ln y is y ' * 1/y:

y is itself a function of x, so ln(y) means ln(y(x)).

y(x) is the inner function and its derivative is y'(x) = dy/dx.

f(z) = ln(z) is the outer function and its derivative is 1/z.

Thus the derivative of f(y(x)) is y'(x) f'(y(x)) = dy/dx * 1/y(x), written just dy/dx * 1/y or y' * 1/y.

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RESPONSE -->

Ok.

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08:09:50

Query problem 3.7.34 (3d edition 3.7.26) was 4.7.18 circle x^2+y^2=25

What are the equations of the tangent lines to the circle at the points where x = 4?

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RESPONSE -->

The circle gets split into the top half and bottom half, each of which have their own equation for the tangent line at x=4. To find this, I first need to find the derivative of x^2 + y^2 = 25:

y' = -x / y

Then I need to find the equation for the top and bottom of the circle to find an (x, y) pair:

top: y = 'sqrt( 25 - x^2 )

bottom: y = - 'sqrt( 25 - x^2)

Solve for at x=4:

top: y = 3, an ordered pair of (4, 3)

bottom: y = -3, an ordered pair of (4, -3)

Then I can take the slope (derivative) of this point:

top: y' = - 4 / 3

bottom: y' = 4 / 3

I can then find the equation of the tangent line at these two points using the point-slope form:

top: y = -(4/3)x + 25/3

bottom: y = (4/3)x - 7/3

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08:17:01

** Solving x^2+y^2=25 for x = 4 you get y^2 = 9, which has solutions y = +3 and y = -3.

So the points are (4, 3) and (4, -3).

By the geometry of the circle the tangent line at a point on the circle is perpendicular to the radial line to that point. The radial lines go from (0,0) to (4,-3) and to (4,3) and so have slopes -3/4 and 3/4, respectively. The tangent lines, being perpendicular, will have the negative reciprocal slopes of 4/3 and -4/3, respectively.

Alternatively implicit differentiation of the equation gives us 2 x + 2 y dy/dx = 0, so that dy/dx = -x/y. At (4,3) and (4,-3) we get dy/dx = -4/3 and +4/3, respectively, confirming the previous results.

Thus, either way, slope at (4,3) is -4/3, slope at (4,-3) is +4/3.

Equation of tangent line at (4,3) is y - 3 = -4/3 ( x - 4) so y = -4/3 x + 25/3.

Equation of tangent line at (4,-3) is y - -3 = 4/3 ( x - 4) so y = 4/3 x - 25/3.

When x = 4 we have 4^2 + y^2 = 25 so y^2 = 25 - 16 = 9 and y = 3 or -3.

Thus the points where x = 4 are (4,3) and (4,-3).

The slope from the center (0,0) to (4,3) is 4/3; the tangent line is perpendicular to the radial line from the center to (4,3), so has slope -4/3. The equation of the tangent line is therefore y - 3 = -4/3 (x-4), or y = -4/3 x + 25/3.

The slope from the center (0,0) to (4,-3) is -4/3; the tangent line is perpendicular to the radial line from the center to (4,-3), so has slope 4/3. The equation of the tangent line is therefore y - (-3) = 4/3 (x-4) or y = 4/3 x - 25/3.

The normal lines are perpendicular to the tangent lines. They therefore have slopes which are the negative reciprocals of the slopes of the tangent lines.

Thus the normal lines pass through (4,3) with slope 3/4 and through (-4/3) with slope -3/4.

You might note that lines through the origin and with the specified slopes pass through the corresponding points, so those lines are y = 3/4 x and y = -3/4 x, respectively.

If you don't notice this you will go ahead and use the point-slope form of the equations. You get

y - 3 = 3/4 ( x - 4), which on solving for y gives you y = 3/4 x, and

y - 3 = -3/4 ( x - (-4)), which on solving for y gives you y = -3/4 x. **

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RESPONSE -->

I see what I missed on the bottom tangent line equation: I didn't correctly reverse the sign of the 3 when I moved it over. Also, that's a neat shortcut to get to the normal line equations.

RIght, but note that this shortcut only works for motion in a plane. In 3 dimensions things get a little more complicated.

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