how to solve this

Question

Consider the curve defined by √𝑥+√𝑦=2.

a) Find 𝑑𝑦/𝑑𝑥.

b) Use your answer from part (a) to find the point where the curve has a tangent line parallel to 𝑦=−2𝑥−7.

Show all your work.

c) Use your answer from part (a) to find the point where the curve has a vertical tangent line. Show all your

work.

(0,4)

Answer 1

The derivative dy/dx=-sqrt(y/x).

The derivative needs to 2 for part b.

The vertical tangent will occur when x=0.

You need to fill in the details...and if you have specific questions, please comment and I will try to help further.

Answer 2

a) √x + √y = 2 → √y = 2 - √x → y = (2 - √x)² = 4 - 4√x + x → y = 4 - 4√x + x

-√x square

Recap: d/dx(c) = 0, c is a constant; d/dx(√x) = 1 / (2√x); d/dx(x) = 1, d/dx(cx) = c*d/dx(x) = c(1) = c

dy/dx = d/dx(4) - d/dx(4√x) + d/dx(x) = 0 - 4 / (2√x) + 1 = -2 / √x + 1

b) By the mean value theorem, there is a point c on the curve where the tangent line at this point is parallel to the slope of the given line y = -2x - 7. In other words, f '(c) = -2.

f '(c) = -2 → -2 / √x + 1 = -2 → -2 / √x = -3 → 2 = 3√x → 2 / 3 = √x

-1 × -√x ÷3 square

The tangent line parallel to the line y = -2x -7 is at x = 4 / 9.

c) Using f '(x) = -2 / √x + 1, we see that the only undefined value of x on the interval (0, 4) is x = 0 which gives a vertical line.