Mastering Parabolas, Trig Identities & Differentiation

Hey math whizzes! Today, we're diving deep into some seriously cool concepts in mathematics. We'll be tackling the general form of a parabola's equation, proving a fundamental trigonometric identity, and getting our hands dirty with some differentiation. So grab your calculators, sharpen your pencils, and let's get this math party started!

Unveiling the Parabola: Vertex and Focus Secrets

Alright guys, let's kick things off with parabolas. You know, those beautiful U-shaped curves that pop up everywhere from physics trajectories to the shape of satellite dishes. Today, we're going to find the general form of the equation of a parabola with a specific vertex and focus. Imagine you've got a parabola where the vertex is chilling at $(2,1)$ and its focus is located at $(2,4)$. What does this tell us, right? The vertex is like the turning point, the absolute lowest or highest point of the parabola. The focus, on the other hand, is a special point that, along with the directrix (which we'll touch on later), defines the parabola. The key thing to notice here is that the x-coordinates of the vertex and the focus are the same (both are 2). This means our parabola is going to open either straight up or straight down, along a vertical line. If the x-coordinates were different, it would open sideways. Since the focus $(2,4)$ is above the vertex $(2,1)$, our parabola is definitely opening upwards. The distance between the vertex and the focus is super important; it's called the focal length, often denoted by '$a$' or '$p$'. In our case, the distance is $4 - 1 = 3$. So, $a=3$. The standard equation for a parabola opening upwards with vertex $(h, k)$ is $(x-h)^2 = 4a(y-k)$. Now, let's plug in our specific values: $h=2$, $k=1$, and $a=3$. This gives us $(x-2)^2 = 4(3)(y-1)$, which simplifies to $(x-2)^2 = 12(y-1)$. This is the standard form. To get the general form, we just need to expand and rearrange it. First, expand $(x-2)^2$ to get $x^2 - 4x + 4$. Then, distribute the 12 on the right side: $12(y-1) = 12y - 12$. So now we have $x^2 - 4x + 4 = 12y - 12$. To put it in the general form $Ax^2 + Bx + Cy + D = 0$ or $y = Ax^2 + Bx + C$, let's aim for the latter since it's already mostly there. We want to isolate $y$. Let's rearrange the equation: $12y = x^2 - 4x + 4 + 12$, which means $12y = x^2 - 4x + 16$. Finally, divide everything by 12 to get $y = \frac{1}{12}x^2 - \frac{4}{12}x + \frac{16}{12}$. Simplifying the fractions, we get $y = \frac{1}{12}x^2 - \frac{1}{3}x + \frac{4}{3}$. And there you have it, the general form of the equation of our parabola! Pretty neat, huh? This equation perfectly describes the path of any point equidistant from the focus $(2,4)$ and the directrix, which would be the line $y = 1 - 3 = -2$ in this case. Understanding the relationship between the vertex, focus, and the resulting equation is fundamental to graphing and analyzing these conic sections.

The Classic Identity: Proving $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$

Next up, let's flex those trigonometric muscles and prove a super famous identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$. This identity holds true for all values of $x$ in the domain $[-1, 1]$. This might seem like magic, but it's actually rooted in the fundamental definitions and properties of inverse trigonometric functions. Let's define $y = \sin^{-1} x$. By definition, this means $\sin y = x$, and the range of $y$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Now, we know a cool trigonometric identity: $\cos(\frac{\pi}{2} - \theta) = \sin \theta$. Let's apply this here. If we let $\theta = y$, then we have $\cos(\frac{\pi}{2} - y) = \sin y$. Since we know $\sin y = x$, we can substitute that in: $\cos(\frac{\pi}{2} - y) = x$. Now, remember our goal is to relate $\sin^{-1} x$ and $\cos^{-1} x$. We have an equation involving $x$ and a cosine function. To isolate the angle, we can take the inverse cosine (arccos) of both sides: $\cos^{-1}(\cos(\frac{\pi}{2} - y)) = \cos^{-1} x$. The inverse cosine function and the cosine function cancel each other out, but we need to be careful about the domain and range. The property $\cos^{-1}(\cos z) = z$ is true when $z$ is in the interval $[0, \pi]$. Let's check if our angle $(\frac{\pi}{2} - y)$ falls within this range. We know that $y$ is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Multiplying by -1 reverses the inequality: $-y$ is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Adding $\frac{\pi}{2}$ to all parts of the inequality: $\frac{\pi}{2} - y$ is in $[-\frac{\pi}{2} + \frac{\pi}{2}, \frac{\pi}{2} + \frac{\pi}{2}]$, which simplifies to $[0, \pi]$. Perfect! Our angle $(\frac{\pi}{2} - y)$ is indeed within the valid range $[0, \pi]$. Therefore, we can confidently say that $\frac{\pi}{2} - y = \cos^{-1} x$. Now, we just need to rearrange this equation to solve for $y$. Add $y$ to both sides: $\frac{\pi}{2} = \cos^{-1} x + y$. And since we originally defined $y = \sin^{-1} x$, we can substitute that back in: $\frac{\pi}{2} = \cos^{-1} x + \sin^{-1} x$. Ta-da! We've successfully proven the identity. This identity is super useful in calculus, especially when dealing with integrals and derivatives involving inverse trigonometric functions. It simplifies many expressions and can save you a lot of work. It highlights the beautiful symmetry and relationship between the sine and cosine functions and their inverses.

Derivative Detectives: Tackling Logarithmic Functions

Finally, let's dive into the world of calculus and sharpen our differentiation skills. We're asked to differentiate two functions with respect to $x$. Get ready, because this involves logarithms, and sometimes those can be a bit tricky, but we'll break it down. First up, we have the function (a) $\ln x \log_{10} x$. This is a product of two functions involving logarithms. To differentiate a product, we absolutely need to use the product rule. Remember the product rule? If you have a function $f(x) = u(x)v(x)$, then its derivative is $f'(x) = u'(x)v(x) + u(x)v'(x)$. Here, let $u(x) = \ln x$ and $v(x) = \log_{10} x$. First, we need the derivatives of $u(x)$ and $v(x)$. The derivative of $\ln x$ is pretty straightforward: $u'(x) = \frac{1}{x}$. Now, for $v(x) = \log_{10} x$, we need to remember how to differentiate logarithms with bases other than $e$. We can use the change of base formula for logarithms, which states $\log_b a = \frac{\ln a}{\ln b}$. So, $\log_{10} x = \frac{\ln x}{\ln 10}$. The term $\ln 10$ is just a constant. Therefore, $v(x) = \frac{1}{\ln 10} \ln x$. Now we can find the derivative of $v(x)$: $v'(x) = \frac{1}{\ln 10} \cdot \frac{1}{x} = \frac{1}{x \ln 10}$. Awesome! We have all the pieces for the product rule. Let's put them together: $f'(x) = u'(x)v(x) + u(x)v'(x) = (\frac{1}{x})(\log_{10} x) + (\ln x)(\frac{1}{x \ln 10})$. We can simplify this a bit. Let's rewrite $\log_{10} x$ using the change of base formula again to get everything in terms of natural logarithms, which often makes things cleaner: $f'(x) = \frac{1}{x} \left(\frac{\ln x}{\ln 10}\right) + (\ln x)(\frac{1}{x \ln 10})$. Notice that both terms have $\frac{\ln x}{x \ln 10}$ in common. So, we can factor that out: $f'(x) = \frac{\ln x}{x \ln 10} + \frac{\ln x}{x \ln 10}$. Wait, that's not right. Let's reapply the product rule carefully. $f'(x) = u'(x)v(x) + u(x)v'(x) = (\frac{1}{x})(\log_{10} x) + (\ln x)(\frac{1}{x \ln 10})$. Now, substitute $\log_{10} x = \frac{\ln x}{\ln 10}$: $f'(x) = \frac{1}{x} \left(\frac{\ln x}{\ln 10}\right) + (\ln x)(\frac{1}{x \ln 10})$. This looks like it simplifies to $f'(x) = \frac{\ln x}{x \ln 10} + \frac{\ln x}{x \ln 10}$, which is $f'(x) = \frac{2 \ln x}{x \ln 10}$. Ah, no, I made a mistake in the previous step. Let's go back. $f'(x) = (\frac{1}{x})(\log_{10} x) + (\ln x)(\frac{1}{x \ln 10})$. Let's simplify the first term using the change of base for $\log_{10} x$: $f'(x) = \frac{1}{x} \left(\frac{\ln x}{\ln 10}\right) + \frac{\ln x}{x \ln 10}$. So, $f'(x) = \frac{\ln x}{x \ln 10} + \frac{\ln x}{x \ln 10}$. This is still leading to the same incorrect simplification. Let's try rewriting the original expression first. $f(x) = \ln x \cdot \frac{\ln x}{\ln 10} = \frac{(\ln x)^2}{\ln 10}$. Now, differentiating this is much easier! Since $\frac{1}{\ln 10}$ is a constant, we can pull it out: $f'(x) = \frac{1}{\ln 10} \cdot \frac{d}{dx}[(\ln x)^2]$. To differentiate $(\ln x)^2$, we use the chain rule. Let $w = \ln x$. Then we have $w^2$. The derivative of $w^2$ with respect to $w$ is $2w$. The derivative of $w = \ln x$ with respect to $x$ is $\frac{1}{x}$. So, by the chain rule, $\frac{d}{dx}[(\ln x)^2] = 2w \cdot \frac{1}{x} = 2(\ln x) \cdot \frac{1}{x} = \frac{2 \ln x}{x}$. Putting it all together: $f'(x) = \frac{1}{\ln 10} \cdot \frac{2 \ln x}{x} = \frac{2 \ln x}{x \ln 10}$. Okay, that's much better and makes more sense. The derivative of $\ln x \log_{10} x$ is $\frac{2 \ln x}{x \ln 10}$.

Now for the second function: (b) This part seems to be missing from the prompt. If there was a second function to differentiate, please provide it, and we'll tackle it with the same enthusiasm!

So there you have it, folks! We've navigated the twists and turns of parabolas, unraveled a fundamental trigonometric truth, and put our differentiation skills to the test. Math can be challenging, but breaking it down step-by-step and understanding the underlying principles makes it totally conquerable. Keep practicing, keep exploring, and don't be afraid to ask questions. Until next time, happy calculating!