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Stewarts - Stewarts is a small group of off-licences in Northern Ireland, that was part of 'Stewarts Supermarket Limited' until the supermarket chain was purchased by Tesco in 1997. Stewarts and Crazy Prices, already owned by the same parent company, merged in Northern Ireland in the 1990s. The company..
Kimberly Stewart - Kimberly Alana Stewart , also known as Kimberley Stewart or Kim Stewart (born 21 August 1979, Holmby Hills, California), is an American socialite, fashion model and fashion designer and the daughter of Rod Stewart. She is the daughter of rock star Rod Stewart and his first wife..
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Fundamental Theorem of Calculus
The fundamental theorem of calculus is the statement that the two central operations of calculus, differentiation and integration, are inverse operations: if a continuous function is first integrated and then differentiated, the original function is retrieve..
First Fundamental Theorem of Integral Calculus
Let f(x) be a continuous function on the closed interval [a, b]. Let the area function A(x) be defined by th..
Let f(x) be a continuous function on the closed interval [a, b]. Let the area function A(x) be defined by th..Second Fundamental Theorem of Integral Calculus
Let f(x) be a continuous function defined on an interval [a,b]. between the limits a and b. This statement is also known as 'fundamental theorem of calculus'. We call b, the upper limit of x and a, the lower limit. If in place of F(x) we take F(x)+c as the value of the integral, we have =..
Let f(x) be a continuous function defined on an interval [a,b]. between the limits a and b. This statement is also known as 'fundamental theorem of calculus'. We call b, the upper limit of x and a, the lower limit. If in place of F(x) we take F(x)+c as the value of the integral, we have =..Solved Problems
Solving equations involving exponentials and logarithms 1. Find x when ln (x + 1) = 5 Solution: x + 1 = e 5 ; x = e 5 - 1 = 147.41 2. Solve 5 x = 7 Solution : Taking natural logarithms both sides, x ln 5 = ln 7. Hence x = ln 7/ln 5 = 1.21 3. Find the value of ..
Are there any free study guides or problem sets for Stewart's Calculus textbook?Our calculus study session?
Where can I find solutions manual to Multivariable Calculus by Stewart with the EVEN numbered problems?Sphere Volume Solution Calculus
Question : does anybody have the complete solutions guide to stewart's calculus 5th edition? i need help with #68 on page 715
or can you help me solve the problem?
determine the points on the curve where the tangent line is horizontal or vertical.
r = cos + sin
thanks in advance :)
Answer : First we convert to cartesian equation using substitutions: x = r cos( ) y = r sin( ) r = x + y r = cos( ) + sin( ) r = r cos( ) + r sin( ) x + y = x + y Now we use implicit differentiation to find dy/dx: 2x + 2y dy/dx = 1 + dy/dx 2y dy/dx - dy/dx = 1 - 2x dy/dx (2y - 1) = 1 - 2x dy/dx = (1 - 2x) / (2y - 1) Tangent is horizontal when dy/dx = 0 (1 - 2x) / (2y - 1) = 0 1 - 2x = 0 x = 1/2 Tangent is vertical when dx/dy = 0 (2y - 1) / (1 - 2x) = 0 2y - 1 = 0 y = 1/2 Now substituting x = 1/2 into equation (x + y = x + y), we can find y: 1/4 + y = 1/2 + y y - y - 1/4 = 0 y = [1 (1+1)] / 2 y = (1 2) / 2 Similarly, substituting y = 1/2 into equation, we get x = (1 2) / 2 Horizontal tangents at points: (1/2, (1+ 2)/2) and (1/2, (1- 2)/2) Vertical tangents at points: ((1+ 2)/2, 1/2) and ((1- 2)/2, 1/2) Now since x = r cos( ) and y = r sin( ), then y/x = r cos( ) / r sin( ) = cos( ) / sin( ) =..
Answer : First we convert to cartesian equation using substitutions: x = r cos( ) y = r sin( ) r = x + y r = cos( ) + sin( ) r = r cos( ) + r sin( ) x + y = x + y Now we use implicit differentiation to find dy/dx: 2x + 2y dy/dx = 1 + dy/dx 2y dy/dx - dy/dx = 1 - 2x dy/dx (2y - 1) = 1 - 2x dy/dx = (1 - 2x) / (2y - 1) Tangent is horizontal when dy/dx = 0 (1 - 2x) / (2y - 1) = 0 1 - 2x = 0 x = 1/2 Tangent is vertical when dx/dy = 0 (2y - 1) / (1 - 2x) = 0 2y - 1 = 0 y = 1/2 Now substituting x = 1/2 into equation (x + y = x + y), we can find y: 1/4 + y = 1/2 + y y - y - 1/4 = 0 y = [1 (1+1)] / 2 y = (1 2) / 2 Similarly, substituting y = 1/2 into equation, we get x = (1 2) / 2 Horizontal tangents at points: (1/2, (1+ 2)/2) and (1/2, (1- 2)/2) Vertical tangents at points: ((1+ 2)/2, 1/2) and ((1- 2)/2, 1/2) Now since x = r cos( ) and y = r sin( ), then y/x = r cos( ) / r sin( ) = cos( ) / sin( ) =..
Question : does anybody have the complete solutions guide to stewart's calculus 5th edition? i need help with #68 on page 715
or can you help me solve the problem?
show that the curves intersect at right angles.
r = a sin , r = a cos
thanks in advance :)
Answer : The point of intersection of the curves is given by a sin = a cos sin = cos = pi/4 For the first curve, r = a sin x = r cos = a sin cos dx/d = a [sin (-sin ) + cos .cos ] dx/d = a(cos^2 - sin^2 ) = a cos(2 ) y = r sin = a sin sin = a sin^2 dy/d = 2a sin cos = a sin(2 ) dy/dx = dy/d / dx/d = [a sin(2 )] / [a cos(2 )] dy/dx = tan(2 ) At the point of intersection, dy/dx = tan(2pi/4) = tan(pi/2) = infinity Therefore the tangent to the first curve is vertical at the point of intersection. _____________________________. For the second curve, r = a cos x = r cos = a cos cos = a cos^2 dx/d = a 2 cos (-sin ) dx/d = -a sin(2 ) y = r sin = a cos sin dy/d = a [cos cos + sin (-sin )] dy/d = a[cos^2 - sin^2 ] dy/d = a cos(2 ) dy/dx = dy/d / dx/d = [a cos(2 )]/[-a sin(2 )] dy/dx = - cot(2 ) At the point of intersection, dy/dx = -cot(2pi/4) = -co..
Answer : The point of intersection of the curves is given by a sin = a cos sin = cos = pi/4 For the first curve, r = a sin x = r cos = a sin cos dx/d = a [sin (-sin ) + cos .cos ] dx/d = a(cos^2 - sin^2 ) = a cos(2 ) y = r sin = a sin sin = a sin^2 dy/d = 2a sin cos = a sin(2 ) dy/dx = dy/d / dx/d = [a sin(2 )] / [a cos(2 )] dy/dx = tan(2 ) At the point of intersection, dy/dx = tan(2pi/4) = tan(pi/2) = infinity Therefore the tangent to the first curve is vertical at the point of intersection. _____________________________. For the second curve, r = a cos x = r cos = a cos cos = a cos^2 dx/d = a 2 cos (-sin ) dx/d = -a sin(2 ) y = r sin = a cos sin dy/d = a [cos cos + sin (-sin )] dy/d = a[cos^2 - sin^2 ] dy/d = a cos(2 ) dy/dx = dy/d / dx/d = [a cos(2 )]/[-a sin(2 )] dy/dx = - cot(2 ) At the point of intersection, dy/dx = -cot(2pi/4) = -co..