Find the gradient of the curve xy^3 = 5 ln y at the point (0,1).
The gradient of the curve at the point (0,1) is 0.
Well, let's put on our clown noses and tackle this one! To find the gradient of the curve xy^3 = 5 ln y at the point (0,1), we'll need to use a little calculus.
First, let's take the derivative of the equation with respect to x. Using the product and chain rule, we have:
y^3 + 3xy^2(dy/dx) = 0
Next, let's solve for dy/dx. We can rearrange the equation to isolate dy/dx:
dy/dx = -y^3 / (3xy^2)
Now substituting x = 0 and y = 1 into the equation, we get:
dy/dx = -1 / (3 * 0 * 1^2)
And, well, dividing by zero isn't exactly our strong suit. So, we can't define the gradient at the point (0,1).
But look on the bright side! At least our clown noses are still intact. Keep those smiles going! 🤡
To find the gradient of the curve xy^3 = 5 ln y at the point (0,1), we need to compute the derivative dy/dx and then substitute x=0 and y=1 into the derivative.
Step 1: Differentiate both sides of the equation xy^3 = 5 ln y with respect to x using the product rule.
Using the product rule, we get:
d(xy^3)/dx = d(5 ln y)/dx.
Differentiating the left side of the equation:
d(xy^3)/dx = y^3 * dx/dx + 3xy^2 * dy/dx.
Differentiating the right side of the equation:
d(5 ln y)/dx = 0.
Simplifying the equation, we have:
y^3 + 3xy^2 * dy/dx = 0.
Step 2: Solve the equation for dy/dx.
To isolate dy/dx, we need to move the terms not involving dy/dx to the other side:
3xy^2 * dy/dx = -y^3.
Now, divide both sides by 3xy^2:
dy/dx = -y^3 / (3xy^2).
Simplifying further, we get:
dy/dx = -1 / (3x).
Step 3: Substitute x=0 and y=1 into the derivative to find the gradient at the point (0,1).
Substituting x=0 and y=1 into the derivative dy/dx:
dy/dx = -1 / (3 * 0) = undefined.
Therefore, the gradient of the curve xy^3 = 5 ln y at the point (0,1) is undefined.
To find the gradient of the curve at the point (0,1), we first need to take the derivative of the equation with respect to x. Let's go through the steps together:
1. Start with the given equation: xy^3 = 5 ln y.
2. Take the derivative of both sides of the equation with respect to x. Recall that when differentiating a product of two functions, we can use the product rule.
So, let's apply the product rule:
Differentiate the left side:
d/dx (xy^3) = x * d/dx(y^3) + y^3 * d/dx(x)
Differentiate the right side:
d/dx (5 ln y) = 5 * d/dx(ln y)
3. Simplify the derivatives:
d/dx(y^3) = 3y^2 * dy/dx (by the power rule)
d/dx(x) = 1 (since the derivative of x with respect to x is 1)
d/dx(ln y) = 1/y * dy/dx (by the chain rule)
4. Substitute the derivatives back into the equation:
x * 3y^2 * dy/dx + y^3 = 5 * 1/y * dy/dx
5. Solve for dy/dx:
Rearrange the equation:
3xy^2 * dy/dx - 5/y * dy/dx = -y^3
Factor out dy/dx:
(3xy^2 - 5/y) * dy/dx = -y^3
Divide both sides by (3xy^2 - 5/y):
dy/dx = -y^3 / (3xy^2 - 5/y)
6. Now that we have the derivative, let's evaluate it at the point (0, 1) to find the gradient:
Substitute x = 0 and y = 1 into dy/dx:
dy/dx = -1^3 / (3(0)(1)^2 - 5/1)
dy/dx = -1 / (-5)
Simplify the expression:
dy/dx = 1/5
Therefore, the gradient of the curve at the point (0,1) is 1/5.