Linear approximation:

Consider the curve defined by
-8x^2 + 5xy + y^3 = -149
a. find dy/dx
b. write an equation for the tangent line to the curve at the point (4,-1)
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 the value of k
d. Write an equation that can be solved to find the actual value of k so that the point (4.2,k) is on the curve
e. Solve the equation found in part (d) for the value of k

I have already figured out:
a. dy/dx=16x-5y/5x +3y^2
b. y+1=3(x-4)

I am just having trouble solving c, d, and e

I would really appreciate any help I can get! Thank you!!

Well, I'm glad you've figured out parts a and b! Now let's move on to parts c, d, and e.

c. To approximate the value of k, we will use the tangent line equation from part b, which is y + 1 = 3(x - 4). We want to find the value of k when x = 4.2. So, substitute x = 4.2 into the equation and solve for y:

y + 1 = 3(4.2 - 4)
y + 1 = 3(0.2)
y + 1 = 0.6
y = -0.4

Therefore, the point (4.2, -0.4) is on the tangent line.

d. Now we need to find the actual value of k so that the point (4.2, k) is on the curve. We can use the given equation -8x^2 + 5xy + y^3 = -149.

Replace x with 4.2 and y with k:

-8(4.2)^2 + 5(4.2)(k) + k^3 = -149

e. Finally, we solve the equation from part d for the value of k. Unfortunately, this equation is not easily solvable by hand as it is a cubic equation. You can use numerical methods like Newton's method or use a graphing calculator or software to approximate the value of k.

c. To approximate the value of k, we will use the equation of the tangent line found in part (b): y + 1 = 3(x - 4).

Substitute x = 4.2 into the equation to find y:
y + 1 = 3(4.2 - 4)
y + 1 = 3(0.2)
y + 1 = 0.6

Solve for y:
y = 0.6 - 1
y = -0.4

Therefore, the approximate value of k is -0.4.

d. To find the actual value of k so that the point (4.2, k) is on the curve, we substitute x = 4.2 and y = k into the equation of the curve:

-8(4.2)^2 + 5(4.2)(k) + (k)^3 = -149

Simplify the equation:
-71.68 + 21k + k^3 = -149

e. To solve the equation -71.68 + 21k + k^3 = -149 for the value of k, rearrange the equation:

k^3 + 21k - 77.32 = 0

This is a cubic equation that can be solved numerically using methods such as graphing calculators or numerical approximation methods.

Please let me know if you need further assistance!

To solve parts (c), (d), and (e), we need to use the tangent line equation obtained in part (b), which is:

y + 1 = 3(x - 4).

c. To approximate the value of k at the point (4.2, k), we can use the tangent line found in part (b). Since the point (4.2, k) is very close to the point (4, -1) on the curve, we can use the tangent line as an approximation. Substituting x = 4.2 in the equation of the tangent line, we get:

k + 1 = 3(4.2 - 4)
k + 1 = 3(0.2)
k + 1 = 0.6

Therefore, the approximate value of k is 0.6.

d. To find the actual value of k, we need to substitute x = 4.2 and y = k into the curve equation and solve for k. Substituting these values into the curve equation, we get:

-8(4.2)^2 + 5(4.2)(k) + k^3 = -149

Simplifying the equation, we have:

-70.56 + 21k + k^3 = -149

e. Rearranging the equation found in part (d), we get:

k^3 + 21k - 70.56 = -149

To solve this equation for the value of k, we can use numerical approximation methods such as the Newton-Raphson method, or we can use a graphing calculator or software to find the approximate value of k.

(c),(d) just plug in 4.2 for x and solve for y. That will be the value for k.

Since (d) will be a cubic in y, it might take a graphical solution.