Consider the curve -8x^2+5xy+y^3=-149
1. Find Dy/dx
2. write an equation for the line tangent to the curve at the point (4,-1)
1. Dy/dx = (-16x + 5y + 3y^2)/(5x + 3y^2)
2. y = -16x + 5(4) - 1 = -16x + 17
To find the derivative, Dy/dx, of the given curve, we need to implicitly differentiate the equation with respect to x.
1. Find Dy/dx:
Start by differentiating both sides of the equation -8x^2 + 5xy + y^3 = -149 with respect to x. The derivative of y^3 with respect to x is 3y^2 * Dy/Dx, and the derivative of xy with respect to x is y + x * Dy/Dx. The derivative of -149 with respect to x is 0.
-16x + 5 * (y + x * Dy/Dx) + 3y^2 * Dy/Dx = 0
Now we can simplify the equation and collect the terms with Dy/Dx on one side:
-16x + 5y + 5x * Dy/Dx + 3y^2 * Dy/Dx = 0
5x * Dy/Dx + 3y^2 * Dy/Dx = 16x - 5y
Dy/Dx * (5x + 3y^2) = 16x - 5y
Dy/Dx = (16x - 5y) / (5x + 3y^2)
Therefore, the derivative Dy/dx of the curve -8x^2 + 5xy + y^3 = -149 is (16x - 5y) / (5x + 3y^2).
2. Write an equation for the line tangent to the curve at the point (4, -1):
To find the equation of the tangent line at a specific point on the curve, we substitute the given point's coordinates into the derivative equation and solve for the slope, m.
Using the coordinates (4, -1), we have:
Dy/dx = (16(4) - 5(-1)) / (5(4) + 3(-1)^2)
= (64 + 5) / (20 + 3)
= 69 / 23
= 3
The slope of the tangent line is 3.
To find the equation of the line, plug in the slope and the coordinates (4, -1) into the point-slope form equation (y - y1) = m(x - x1):
(y - (-1)) = 3(x - 4)
y + 1 = 3x - 12
y = 3x - 13
Therefore, the equation of the line tangent to the curve -8x^2 + 5xy + y^3 = -149 at the point (4, -1) is y = 3x - 13.
To find Dy/dx, we need to take the derivative of the equation with respect to x. This can be done by applying the chain rule and product rule as necessary.
1. Find Dy/dx:
Given equation: -8x^2 + 5xy + y^3 = -149
To take the derivative of this equation with respect to x, we need to treat y as a function of x. Let's denote y as f(x) for simplicity.
Differentiating both sides of the equation with respect to x, we get:
d/dx(-8x^2 + 5xf(x) + f(x)^3) = d/dx(-149)
Using the product rule for the second term, we have:
-16x + 5f(x) + 5xf'(x) + 3f(x)^2 * f'(x) = 0
Now, let's solve this equation for f'(x) (which represents Dy/dx):
5xf'(x) + 3f(x)^2 * f'(x) = 16x - 5f(x) - 3f(x)^2
Factoring out f'(x) on the left side:
f'(x) * (5x + 3f(x)^2) = 16x - 5f(x) - 3f(x)^2
Dividing both sides by (5x + 3f(x)^2):
f'(x) = (16x - 5f(x) - 3f(x)^2) / (5x + 3f(x)^2)
So, the derivative Dy/dx is given by:
Dy/dx = (16x - 5y - 3y^2) / (5x + 3y^2)
2. To write the equation for the line tangent to the curve at the point (4,-1), we need to find both the slope (m) and the point (x0, y0) on the curve.
We already know the slope, which is the derivative Dy/dx evaluated at the point (4,-1).
Using the equation for Dy/dx we obtained in step 1, substitute x = 4 and y = -1:
Dy/dx = (16(4) - 5(-1) - 3(-1)^2) / (5(4) + 3(-1)^2)
= (64 + 5 + 3) / (20 + 3)
= 72 / 23
So, the slope (m) is 72/23.
Now, to find the point (x0, y0) on the curve, substitute x = 4 into the equation:
-8(4^2) + 5(4)(y) + y^3 = -149
Simplifying:
-128 + 20y + y^3 = -149
Rearranging:
y^3 + 20y = -21
By guessing or by using numerical methods, we find that one possible solution is y = -1.
So, the point (x0, y0) on the curve is (4, -1).
Now, using the point-slope form of a line equation, we can write the equation of the tangent line:
y - y0 = m(x - x0)
Substituting the values we found:
y - (-1) = (72/23)(x - 4)
y + 1 = (72/23)(x - 4)
Now, we can simplify this equation, if necessary.