#$&* course MTH 173 6/9/2013 12:57AM 020. `query 20
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Given Solution: `a** This is a composite with f(z) = e^z and g(x) = ln(x) + 1. g'(x) = 1/x, f'(z) = e^z. So the derivative is (e^(ln(x)+1)) ' = (f(g(x)) ' = g'(x) * f'(g(x)) = 1/x e^(ln(x)+1)) = 1/x e^(ln(x)) * e^1 = 1/x * x * e = e. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhy is in easier to calculate the derivative of this function if you simplify the function first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: e^(ln x + 1) = e^(ln x) * e^1 = x * e or just e * x. e is considered a constant (Euler’s number) so the derivative of e * x is e * 1 = e. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Tangent line at x = 1 passes through the corresponding graph point (1, ln(1) ) = (1, 0). Slope of tangent line equals derivative: y' = 1/x; at x = 1, y' = 1 is obtained The line has slope 1 and passes through (1, 0) The equation of the line is y - 0 = 1 * (x - 1), or y = x-1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** tan line at x = 1 passes through the corresponding graph point (1, ln(1) ) = (1, 0). Slope of tan line equals derivative: y' = 1/x; at x = 1 we have y' = 1. So line has slope 1 and passes through (1, 0). The equation of the line is y - 0 = 1 * (x - 1), or y = x-1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat are your approximations to ln(1.1) and ln(2), based on the tangent line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln(1.1): y = 1.1 - 1 = .1 approx ln(2): y = 2 - 1 = 1 ln(1.1)=0.095 ln(2)=0.69 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** to get approx ln(1.1): y = 1.1 - 1 = .1. to get approx ln(2): y = 2 - 1 = 1. Actual values are ln(1.1)=0.095 and ln(2)=0.69. Note that the first is a little below the approximation given by the tangent line, the second much below the tangent-line value. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: In terms of the graph, explain whether your approximations are larger or smaller than the true values. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph of y = ln(x) is increasing and concave downward and will always curve progressively away from any tangent-line approximation. This makes the approximating point x = 1. The further x is from the point, the further the tangent line will be from the actual graph. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: ** Since the graph of y = ln(x) is increasing and concave downward, it will always curve progressively away from any tangent-line approximation. In this case the approximating point is x = 1. The further x is from the approximating point, the further the tangent line will be from the actual graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.7.16 (was 3.7.9 was 4.7.6) e^(x^2) + ln y = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative of the equation is 2x e^(x^2) + 1/y * y ' = 0. 1/y * y ' = 2 x e^(x^2) y ' = - 2x e^(x^2) * y The derivative of ln y is y ' * 1/y and y is a function of x, so ln(y) means ln(y(x)) y(x) is the inner function and its derivative is y'(x) = dy/dx. f(z) = ln(z) is the outer function and its derivative is 1/z. The derivative of f(y(x)) is y'(x) f ' (y(x)) = dy/dx * 1 / y(x), written as dy/dx * 1/y or y' * 1/y. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aWhen 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. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Solving x^2+y^2=25 for x = 4 you get y^2 = 9, which has solutions y = +3 and y = -3. The points are (4, 3) and (4, -3). The slope at (4,3) is -4/3 and the slope at (4,-3) is +4/3. The tangent line equation at (4,3) is y - 3 = -4/3 ( x - 4) so y = -4/3 x + 25/3. The tangent line equation at (4,-3) is y - -3 = 4/3 ( x - 4) so y = 4/3 x - 25/3. When x = 4 4^2 + y^2 = 25 so y^2 = 25 - 16 = 9 and y = 3 or -3 where x = 4 are (4,3) and (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 y - 3 = -4/3 (x-4), or y = -4/3 x + 25/3. 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 y - (-3) = 4/3 (x-4) or y = 4/3 x - 25/3. Thus the normal lines pass through (4,3) with slope 3/4 and through (-4/3) with slope -3/4. y - 3 = 3/4 ( x - 4) y = 3/4 x y - 3 = -3/4 ( x - (-4)) y = -3/4 x. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!