Question

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$\left(x^{2}+y^{2}\right)^{2}=\frac{25}{4} x y^{2} ;(1,2)$$ (Graph cant copy)

Short answer

Answer: The equation of the tangent line at point (1,2) is $$y = 3x - 1$$.

Step-by-step solution

Verify if the point lies on the curve

Plug the point (1,2) into the given equation and check if it holds true. $$\left(1^{2}+2^{2}\right)^{2}=\frac{25}{4}(1)(2)^{2}$$ $$\left(1+4\right)^{2}=25(1)$$ $$5^{2}=25$$ $$25=25$$ Since the equation holds true, we now know that the point (1,2) lies on the curve.

Find the partial derivatives

We will find the partial derivatives of the given equation concerning x and y. Let $$F(x,y)=\left(x^{2}+y^{2}\right)^{2}-\frac{25}{4}xy^{2}$$, then we need to find $$\frac{\partial F}{\partial x}$$ and $$\frac{\partial F}{\partial y}$$. $$\frac{\partial F}{\partial x} = 4(x^{2}+y^{2})(2x) - \frac{25}{4}y^{2}$$ $$\frac{\partial F}{\partial y} = 4(x^{2}+y^{2})(2y) - \frac{25}{4}x(2y)$$

Evaluate the gradient of the function at point (1,2)

Now we need to evaluate the gradient vector of the function at point (1,2) $$\left.\frac{\partial F}{\partial x}\right|_{(1,2)} = 4(1^{2}+2^{2})(2\cdot1) - \frac{25}{4}\cdot2^{2} = 20 - 25 = -5$$ $$\left.\frac{\partial F}{\partial y}\right|_{(1,2)} = 4(1^{2}+2^{2})(2\cdot2) - \frac{25}{4}\cdot1\cdot(2\cdot2) = 40 - 25 = 15$$ The gradient of the function at point (1,2) is (-5, 15).

Find the equation of the tangent line

Use the gradient and the given point (1,2) to find the equation of the tangent line using the point-slope form. The slope of our tangent line is equal to the ratio $$\frac{dy}{dx}$$. So, $$\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial x}} = -\frac{15}{-5} = 3$$. Using point-slope form, $$y - y_{1} = m(x - x_{1})$$, where m is the slope and (x1, y1) is the point (1,2). Plug in the values: $$y - 2 = 3(x - 1)$$ Finally, $$y = 3x - 1$$. The equation of the line tangent to the curve at the given point (1,2) is $$y = 3x - 1$$.

Key concepts

Curve

In mathematics, a curve can be thought of as a line that smoothly changes direction, and it's described by a function or an equation in both two-dimensional and three-dimensional spaces. For this particular exercise, we are interested in the curve defined by the equation \((x^2 + y^2)^2 = \frac{25}{4}xy^2\).
This equation represents a relationship between the variables \(x\) and \(y\), allowing us to visualize the path these points create when plotted.

• Curves can be straight, like a line, or bent in various shapes, like circles and ellipses.
• The exact form of a curve depends on the equation defining it.

When we need to analyze a curve, like in this exercise, verifying that a point lies on it involves substituting the point's coordinates into the equation and seeing if both sides are equal. If they are, the point is indeed on the curve.

Partial Derivatives

Partial derivatives are used to study functions with multiple variables. In the context of this exercise, we have a function \(F(x,y)\) that depends on both \(x\) and \(y\).
When we take the partial derivative of \(F\) with respect to \(x\), written as \(\frac{\partial F}{\partial x}\), we treat \(y\) as if it's a constant and only differentiate with respect to \(x\). The same goes for \(\frac{\partial F}{\partial y}\), but treating \(x\) as a constant.

• Partial derivatives give us the rate at which the function changes when we vary one variable, holding others constant.
• They are crucial in finding the slope of the tangent line to a curve in multidimensional space.

In our exercise, computing these derivatives helps us understand how the curve behaves locally at any given point.

Gradient

The gradient is a vector that contains all the partial derivatives of a function. For a function \(F(x, y)\), the gradient is denoted by \(abla F\) and represented as a vector \((\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y})\).
In this exercise, evaluating the gradient at point (1,2) involves plugging in these coordinates into the partial derivatives we calculated.

• The gradient points in the direction of the steepest ascent in the function's value.
• It shows how much and in which direction the function increases at a specific point.

Understanding the gradient is essential for finding tangent lines, as it provides the necessary "rise over run" or slope information needed for the tangent line's equation at that point.

Point-Slope Form

The point-slope form is a way to create linear equations. It's particularly useful for finding the equation of a line when you know a point on the line and the line’s slope. The formula for point-slope form is given by \(y - y_1 = m(x - x_1)\).
Here, \((x_1, y_1)\) is a known point, and \(m\) is the slope of the line.

• This method simplifies finding the tangent line at a point on a curve, once the slope is known.
• In the exercise, using this form, we identify the slope from the ratio \(\frac{\partial F}{\partial y} / \frac{\partial F}{\partial x} \).

Applying point-slope form in this exercise results in the tangent line equation \(y = 3x - 1\), demonstrating the utility of this method in deriving the tangent line from known data.