Consider the curve defined by -8x^2 + 5xy + y^3 = -149?

Question

a) compute dy/dx
I got (16x-5y)/(5x + y^2)

b)write an equation for the line tangent to the curve at the point (4,-1)
I plugged the numbers in the derivative and then got (y+1)=3(x - 4)

c)there is a number k so that the point (4.2,k) is on the curve. Using the tangent line found in part b, approximate value of k ..
so I dont know if this is right but I just plugged those numbers..
k+1=3(4.2 - 4) and got k= -0.4

d)write an equation that can be solved to find k so that the point (4.2, k) is on the curve..
What do I do for this one?

e) Solve the equation found in d for the value of k.

Can someone help me?
Thanks!

Answer 1

1)
-16x +5(1y+xdy/dx) +3y^2dy/dx = 0
-16x +5y+ 5xdy/dx +3y^2dy/dx = 0
dy/dx(5x+3y^2) = (16x-5y)
dy/dx = (16x-5y)/(5x+3y^2)
i think you missed a 3 before the y^2 in your answer, becuase you have
the rest correct.. so i assume it was a typo
2) since you have dy/dx just substitue the given numbers and find the value of the gradient:
when you do subtitue and evalue the gradient is:
69/23 which is equal to 3
now using
y-y1 = m(x-x1):
y+1 =3(x-4)-> whic you have correct
y = 3x-13 is the equation of the tangent at that point

3) here, since you have everything else you go back to:
y-y1 = m(x-x1), substitute numbers in:
-1-k= 3(4-4.2)
solving for k
k is equal to -2/5 which is -0.4
which you also got
4) this is pretty simple:
just substitue those values into the original equation:
the values being x = 4.3 and y =k, substitue these everytime you see x and y in the original equation like shown below:
-8(4.2)^2 +5(4.2)k +(k)^3 = -149
-141.12+21k +k^3 = -149
use a graphical calculator or solve otherwise
solving this, one obtains:
-0.37 for the value of k
as you can see this is pretty close to the value we approximated above which was -0.4!!

Answer 2

Your either know the answer or the answer is pointless.