Let's solve the problem step by step from the image.

Problem:

We are asked to calculate the tangent and normal for the function $y = x\sqrt{x} + 2$ at the point $(x_0 = 4, y_0 = x_0\sqrt{x_0} + 2)$.

Step 1: Find the equation of the tangent line

To find the equation of the tangent line at the given point, we need the derivative of the function, which gives us the slope of the tangent line.

The function is: $y = x\sqrt{x} + 2 = x \cdot x^{1/2} + 2 = x^{3/2} + 2$

Derivative of the function $y = x^{3/2} + 2$:

$\frac{dy}{dx} = \frac{d}{dx}\left( x^{3/2} + 2 \right) = \frac{3}{2} x^{1/2}$

Slope of the tangent line at $x_0 = 4$:

Substitute $x_0 = 4$ into the derivative: $\frac{dy}{dx}\bigg|_{x=4} = \frac{3}{2} \cdot 4^{1/2} = \frac{3}{2} \cdot 2 = 3$ So, the slope of the tangent line is $m_t = 3$.

Point $(x_0, y_0)$:

Now, we calculate $y_0$ using the original function at $x_0 = 4$: $y_0 = 4\sqrt{4} + 2 = 4 \cdot 2 + 2 = 8 + 2 = 10$ Thus, the point is $(4, 10)$.

Tangent line equation:

Using the point-slope form $y - y_1 = m(x - x_1)$, with $m = 3$ and $(x_1, y_1) = (4, 10)$: $y - 10 = 3(x - 4)$ Simplifying: $y_t = 3x - 12 + 10 \quad \Rightarrow \quad y_t = 3x - 2$

Step 2: Find the equation of the normal line

The slope of the normal line is the negative reciprocal of the tangent slope. Since the slope of the tangent line is $3$, the slope of the normal line is: $m_n = -\frac{1}{3}$

Normal line equation:

Using the point-slope form for the normal line: $y - 10 = -\frac{1}{3}(x - 4)$ Simplifying: $y - 10 = -\frac{1}{3}x + \frac{4}{3}$ $y_n = -\frac{1}{3}x + \frac{4}{3} + 10 = -\frac{1}{3}x + \frac{34}{3}$

Final Answers:

• The algebraic expression for the vertical coordinate in the tangent equation in terms of $x$ is: $y_t = 3x - 2$
• The algebraic expression for the vertical coordinate in the normal equation in terms of $x$ is: $y_n = -\frac{1}{3}x + \frac{34}{3}$

Would you like further details or clarifications on any part of this? Here are 5 questions that build upon this problem:

1. How do we differentiate a composite function like $y = x\sqrt{x}$?
2. What is the geometric significance of the tangent line at a point on a curve?
3. How would you calculate the normal line if the function was more complex?
4. Why is the slope of the normal line the negative reciprocal of the tangent line’s slope?
5. What are some common uses of tangent and normal lines in physics or engineering?

Tip: When working with functions involving roots, rewrite them using exponents for easier differentiation.