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Math Help - gradient

  1. #1
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    gradient

    1.find the gradient to the curve y = x^3 - 3x^2 at the point x = 3
    then find the equations of the tangent and the normal to the curve at the point where x = 3

    2. the curve with equation y= x+(24/x) passes through the points A (4,10) and B(8,11)

    find the equation of the tangent to the curve at A then the normal at B
    thanks
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  2. #2
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    Quote Originally Posted by pop_91 View Post
    1.find the gradient to the curve y = x^3 - 3x^2 at the point x = 3
    then find the equations of the tangent and the normal to the curve at the point where x = 3

    2. the curve with equation y= x+(24/x) passes through the points A (4,10) and B(8,11)

    find the equation of the tangent to the curve at A then the normal at B
    thanks
    These are really just applications of the basic concepts. The tangent line to a curve has equation y= mx+ b where m is the derivative of the equation of the curve. y= x^3- 3x^2 so y'= 3x^2- 6x and, at x= 3, that is 3(9)- 6(3)= 9. The equation is y= 9x+ b. When x= 3, y= 3^3- 3(3)^2= 0 so 0= 9(3)+ b: b= -27 and y= 9x- 27. That's the equation of the tangent line at x= 3. The equation of the normal to the curve is y= (-1/9)x+ b because the product of the slopes of two perpendicular lines is -1. When x= 3, y= 9 still so 9= (-1/9)(3)+ b. b= 9+ 1/3= 28/3. y= (-1/9)x+ 28/3.

    y= x+ (24/x)= x+ 24x^{-1} has derivative y'= 1- 24x^{-2}. At x= 4, that is 1- 24/16= 1- 3/2= -1/2. The tangent line at (4, 10) is y= (-1/2)x+ b for some b. Putting x= 4, y= 10, 10= (-1/2)(4)+ b= -2+ b so b= 12. The tangent line at A is y= (-1/2)x+ 12.
    At x= 8, y'= 1- 24/64= 1- 3/8= 5/8. The normal line is of the form y= (-8/5)x+ b. When x= 8, y= 11 so 11= (-8/5)(4)+ b= b= 11+ 32/5= (55+ 32)/5= 87/5. The normal line at B is y= (-8/5)x+ 87/5.
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