Explanation 1) Concept: Area between curve: The area A of the region bounded by curves y = f x , y = g x and the line x = c , x = d , where f and g are continuous and f ( x ) ≥ g ( x ) for all x in [ c , d ] , is A = ∫ c d f x - g x d x 2) Calculation: Assume P lies on the curve and in first quadrant. As y = x 3 lies in the first and third quadrant, this assumption is valid. Let P = a , a 3 . Use the below information to find the slope of the tangent line to curve y = x 3 at the point P = ( a , a 3 ) Here, f x = x 3 and f a = a 3 , so the slope is m = lim x → a ⁡ f x - f a x - a = lim x → a ⁡ x 3 - a 3 x - a = lim x → a ⁡ ( x - a ) ( x 2 + x a + a 2 ) x - a = lim x → a ⁡ ( x 2 + x a + a 2 ) = 3 a 2 Using point slope form, find the equation of tangent line at a , a 3 . y - y 1 = m ( x - x 1 ) y - a 3 = 3 a 2 ( x - a ) y = 3 a 2 x - 2 a 3 Now plot the curve and tangent line to find the point Q . It is the intersection of the curve y = x 3 and tangent line y = 3 a 2 x - 2 a 3 . To find the point Q : x 3 = 3 a 2 x - 2 a 3 x 3 - 3 a 2 x + 2 a 3 = 0 x - a 2 x + 2 a = 0 So x = a and x = - 2 a Our point P corresponds to the first value x = a So required x = - 2 a , y = - 2 a 3 = - 8 a 3 So point Q is ( - 2 a , - 8 a 3 ) . Now we need to find the slope of the tangent line to the given curve passing through Q . Here, f x = x 3 and f a = - 8 a 3 , so the slope is m = lim x → a ⁡ f x - f a x - a = lim x → - 2 a ⁡ x 3 + 8 a 3 x + 2 a = lim x → - 2 a ⁡ ( x + 2 a ) ( x 2 - 2 x a + 4 a 2 ) x + 2 a = lim x → - 2 a ⁡ ( x 2 - 2 x a + 4 a 2 ) = 12 a 2 Using the point slope form, find the equation of the tangent line at ( - 2 a , - 8 a 3 ) . y - y 1 = m ( x - x 1 ) y + 8 a 3 = 12 a 2 ( x + 2 a ) y = 12 a 2 x + 16 a 3 To find the point P 2 , which is the intersection of this tangent line at Q and the curve y = x 3 . x 3 = 12 a 2 x + 16 a 3 x 3 - 12 a 2 x - 16 a 3 = 0 x + 2 a 2 x - 4 a = 0 So, x = 4 a , y = 4 a 3 = 64 a 3 So point P 2 is ( 4 a , 64 a 3 )

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ISBN: 9781285740621

Author: James Stewart

Publisher: Cengage Learning

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Chapter 5 Solutions

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