Consider a curve given implicitly by the equation (1+x)y^3 + (x^4)y - 85 = 0.
A. Calculate dy/dx at a general point (x,y).
B. Write the equation of the tangent line to the curve at the point (3,1).
C. At (3,1), y(x) is defined implicitly as a function of x. Let g(x) be the inverse function of y(x). Compute g'(1).
A
(1+x)y^3 + (x^4)y - 85 = 0
y^3 + 3(1+x)y^2y' + 4x^3y + x^4y' = 0
y' = -(y^3+4x^3y)/(3(1+x)+x^4)
= -(y^3+4x^3y)/(x^4+3x+3)
B
y'(3) = -109/93
So, the tangent line is
y-1 = -109/93 (x-3)
C
y(3) = 1, so g(1) = 3
g'(1) = 1/y'(3) = -93/109
A. Calculating dy/dx at a general point (x,y) involves taking the derivative of the equation with respect to x. Let's start by differentiating both sides of the equation:
d/dx[(1+x)y^3 + (x^4)y - 85] = 0
Using the product rule and chain rule, we get:
(1+x)(3y^2)(dy/dx) + y^3 + (x^4)(dy/dx) + 4x^3y - 0 = 0
Now we can solve for dy/dx:
(dy/dx)((1+x)(3y^2) + (x^4)) = -y^3 - 4x^3y
dy/dx = (-y^3 - 4x^3y) / ((1+x)(3y^2) + (x^4))
B. To find the equation of the tangent line at the point (3,1), we need to find the derivative dy/dx at that point and then use the point-slope form of the equation of a line.
Substituting x = 3 and y = 1 into the previously derived expression:
dy/dx = (-(1)^3 - 4(3)^3(1)) / ((1+3)(3(1)^2) + (3^4))
dy/dx = (-1 - 4(27))/ (4(3) + 81)
dy/dx = (-1 - 108) / (12 + 81)
dy/dx = -109 / 93
Using the point-slope form:
y - y1 = m(x - x1)
y - 1 = (-109/93)(x - 3)
Simplifying:
y = (-109x/93) + (526/93)
Hence, the equation of the tangent line to the curve at the point (3,1) is y = (-109x/93) + (526/93).
C. We know that g(x) is the inverse function of y(x). To compute g'(1), we can use the fact that the derivative of an inverse function is the reciprocal of the derivative of the original function evaluated at the corresponding point.
Since y(3) = 1, we can find dy/dx at x = 3 using the earlier derived expression:
dy/dx = (-y^3 - 4x^3y) / ((1+x)(3y^2) + (x^4))
dy/dx = (-(1)^3 - 4(3)^3(1)) / ((1+3)(3(1)^2) + (3^4))
dy/dx = (-1 - 108) / (4(3) + 81)
dy/dx = -109 / 93
Now, to find g'(1), we take its reciprocal:
g'(1) = 1 / (dy/dx)
g'(1) = 1 / (-109/93)
g'(1) = -93/109
Therefore, g'(1) = -93/109. Just like a clown driving a banana car backward, it seems we're moving in the opposite direction!
A. To calculate dy/dx at a general point (x, y), we differentiate the equation implicitly with respect to x.
Differentiating (1+x)y^3 + (x^4)y - 85 = 0 with respect to x, we get:
3(1+x)y^2 * dy/dx + y^3 + 4x^3y^2 + x^4 * dy/dx = 0
Rearranging the terms, we have:
dy/dx = -(y^3 + 4x^3y^2) / (3(1+x)y^2 + x^4)
B. To find the equation of the tangent line to the curve at the point (3,1), we need the slope of the tangent line at that point. So, calculate dy/dx by substituting x = 3 and y = 1 into the expression derived in part A:
dy/dx = -(1^3 + 4(3)^3 * 1^2) / (3(1+3)*1^2 + 3^4)
= -(1 + 4 * 27) / (3 * 4 + 81)
= -109 / 93
Now, we have the slope, m = -109/93, and we also have the point (3,1) on the tangent line. Using the point-slope form of the line equation, we can write the equation of the tangent line as:
y - 1 = (-109/93)(x - 3)
Simplifying further, we get:
y = (-109/93)x + 328/93
C. To find g'(1), we first need to express y as a function of x. At (3,1), the point of interest, we have:
(1+3)y^3 + (3^4)y - 85 = 0
4y^3 + 81y - 85 = 0
We can solve this equation numerically or graphically to find y ≈ 0.92895.
Now, we want to find the derivative of the inverse function g(x) at x = 1. Using the inverse function theorem, we have the relationship:
g'(x) = 1 / (dy/dx)
Thus, we need to evaluate dy/dx at (3,1) and take its reciprocal:
dy/dx = -(y^3 + 4x^3y^2) / (3(1+x)y^2 + x^4)
Substituting x = 3 and y ≈ 0.92895 into the expression, we get:
dy/dx ≈ -((0.92895)^3 + 4(3)^3(0.92895)^2) / (3(1+3)(0.92895)^2 + 3^4)
Finally, we can find g'(1) by taking the reciprocal:
g'(1) ≈ 1 / dy/dx
A. To calculate dy/dx at a general point (x,y), we can use implicit differentiation. Here's how:
Step 1: Take the derivative of both sides of the equation with respect to x.
d/dx[(1+x)y^3 + (x^4)y - 85] = d/dx[0]
Step 2: Apply the product rule and chain rule to the terms on the left side.
(1+x)(3y^2(dy/dx)) + y^3 + (x^4)(dy/dx) + 4x^3y - 0 = 0
Step 3: Rearrange the terms and solve for dy/dx.
(1+x)(3y^2(dy/dx)) + (x^4)(dy/dx) = -4x^3y - y^3
(1+x)(3y^2 + x^4)(dy/dx) = -4x^3y - y^3
dy/dx = (-4x^3y - y^3) / ((1+x)(3y^2 + x^4))
Therefore, dy/dx at a general point (x,y) is (-4x^3y - y^3) / ((1+x)(3y^2 + x^4)).
B. To write the equation of the tangent line to the curve at the point (3,1), we need two pieces of information: the derivative (dy/dx) at that point and the point itself.
Step 1: Calculate the derivative dy/dx at (3,1):
Substitute x = 3 and y = 1 into the expression for dy/dx:
dy/dx = (-4(3)^3(1) - (1)^3) / ((1+3)(3(1)^2 + (3)^4))
= (-108 - 1) / (4(3) + 81)
= -109 / 93
= -1.172
Step 2: Use the point-slope form of a line (y - y1) = m(x - x1), where (x1, y1) is the point and m is the slope:
Substitute x1 = 3, y1 = 1, and m = -1.172 into the equation:
y - 1 = -1.172(x - 3)
So, the equation of the tangent line to the curve at the point (3,1) is y = -1.172x + 4.516.
C. We are given that y(x) is defined implicitly as a function of x at (3,1). Let g(x) be the inverse function of y(x). To compute g'(1), we need to find the derivative of the inverse function at the point (1,g(1)).
Step 1: Find g(x) by interchanging x and y in the equation of the curve:
(1+y)x^3 + (y^4)x - 85 = 0
x(1+y)x^2 + x(y^4) - 85 = 0
x^3 + xyx^2 + xy^4 - 85 = 0
x^3 + xy(x^2 + y^3) - 85 = 0
So, g(x) = x^3 + xy(x^2 + y^3) - 85.
Step 2: Find g'(x) by differentiating g(x) with respect to x:
g'(x) = 3x^2 + (1)(x^2 + y^3) + x(x^2 + y^3) + x(4y^3)
Step 3: Find g'(1) by substituting x = 1 and y = g(1) into g'(x):
g'(1) = 3(1)^2 + (1)((1)^2 + (g(1))^3) + (1)((1)^2 + g(1)^3) + (1)(4(g(1))^3)
Therefore, g'(1) = 3 + 1 + 2g(1) + 4(g(1))^3.
Note: To compute g'(1), we need to know the value of g(1) which is not provided in the given information.