Answered

Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Please ASAP
With explanation PLEASE

( a ) For the curve C with equation y = f(x); dy dx =x^ 3 +2x-7. Given that the point P(2, 4) lies on C find an equation for the normal to C at P in the form ax + by + c = 0 , where a , b and c are integers

(b) y = x ^ p + 2x ^ q where is the derived function . dy dx =11x^ 10 +10x^ 4 ,; (dy)/(dx) Find the value of p and the value of q .


Sagot :

Answer:

  (a) x +5y = 22

  (b) p = 11, q = 5

Step-by-step explanation:

(a)

The derivative of a function tells you the slope of its curve at every point. Then the slope of C at x=2 is ...

  dy/dx = 2³ +2(2) -7 = 5

The normal to the curve at the point of interest will have a slope that is the opposite reciprocal of this: -1/5. Then the point-slope equation of the normal line can be written as ...

  y -k = m(x -h) . . . . line with slope m through point (h, k)

  y -4 = -1/5(x -2) . . . . line with slope -1/5 through point (2, 4)

  5y -20 = -x +2 . . . . multiply by 5

  x +5y = 22 . . . . . . add x+20 to put in standard form

The graph shows curve C and the desired normal line.

__

(b)

The power rule for derivatives tells you ...

  (d/dx)(a·x^n) = a·n·x^(n-1)

This relation gives you two ways to find the values of p and q.

  a) using the exponent of the term

  b) using the coefficient of the term

Either way, the values are ...

  p = 10 +1 = 11

  q = 4 +1 = 10/2 = 5

View image sqdancefan
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.