Find Original Function: Derivative & Point Given

Hey math whizzes and calculus adventurers! Today, we're diving deep into the awesome world of integration, specifically tackling a super common problem: finding the original function when you've got its derivative and a specific point it needs to pass through. Think of it like being a detective, but instead of solving a crime, you're uncovering the secret identity of a function! We've got a hot one today: rac{d y}{d x}=x oldsymbol{\sqrt{x^2+1}} and we know it has to pass through the point $(0,-3)$. This is where the magic of antiderivatives comes into play. We're essentially reversing the differentiation process. So, grab your calculators, dust off those integration rules, and let's get this done!

The Undoing Process: Integration

Alright guys, the first step in our detective mission is to find the general antiderivative of the given derivative, rac{d y}{d x}=x oldsymbol{\sqrt{x^2+1}}. This means we need to integrate this expression with respect to $x$. Our goal is to find $y(x)$. So, we're looking at oldsymbol{y = \int x \sqrt{x^2+1} \, dx}. Now, this integral might look a little intimidating with that square root and the $x$ term hanging out front, but don't sweat it! This is a prime candidate for a technique called u-substitution. It's like a secret code that simplifies complex integrals. We want to choose a part of the integrand that, when differentiated, gives us another part of the integrand. Let's try setting oldsymbol{u = x^2+1}.

Why this choice, you ask? Well, if we differentiate $u$ with respect to $x$, we get oldsymbol{rac{du}{dx} = 2x}. This is super handy because we have an $x$ in our original integral! Now, we can rearrange this to get oldsymbol{du = 2x \, dx}. Looking back at our integral oldsymbol{\int x \sqrt{x^2+1} \, dx}, we see we have an $x \, dx$ term. We can get this by dividing our $du$ equation by 2: oldsymbol{rac{1}{2} du = x \, dx}. Perfect!

Now we can substitute! Replace oldsymbol{x^2+1} with oldsymbol{u} and oldsymbol{x \, dx} with oldsymbol{rac{1}{2} du}. Our integral transforms into something much friendlier: oldsymbol{\int \sqrt{u} \cdot \frac{1}{2} du}. We can pull that constant oldsymbol{\frac{1}{2}} right out of the integral: oldsymbol{\frac{1}{2} \int \sqrt{u} \, du}. Remember that oldsymbol{\sqrt{u}} is the same as oldsymbol{u^{\frac{1}{2}}}. So, we have oldsymbol{\frac{1}{2} \int u^{\frac{1}{2}} \, du}.

This is a standard power rule integration, guys! The power rule states that oldsymbol{\int w^n \, dw = \frac{w^{n+1}}{n+1} + C} (where $n \neq -1$). In our case, $w$ is $u$ and $n$ is oldsymbol{\frac{1}{2}}. So, applying the rule, we get: oldsymbol{\frac{1}{2} \cdot \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} + C}. That simplifies to oldsymbol{\frac{1}{2} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C}.

Dividing by oldsymbol{\frac{3}{2}} is the same as multiplying by oldsymbol{\frac{2}{3}}. So, we have oldsymbol{\frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} + C}. The oldsymbol{\frac{1}{2}} and oldsymbol{\frac{2}{3}} cancel out nicely to give us oldsymbol{\frac{1}{3} u^{\frac{3}{2}} + C}.

But wait! We're not done yet. Remember, our original variable was $x$, not $u$. We need to substitute back oldsymbol{u = x^2+1}. So, our general antiderivative is oldsymbol{y = \frac{1}{3} (x^2+1)^{\frac{3}{2}} + C}. This oldsymbol{+ C} is the constant of integration. It's crucial because the derivative of any constant is zero, meaning there are infinitely many functions that have the same derivative. We need our specific point to nail down which one is ours!

Pinpointing the Specific Function: Using the Given Point

Now that we have the general form of our function, oldsymbol{y = \frac{1}{3} (x^2+1)^{\frac{3}{2}} + C}, it's time to use that secret clue: the point $(0,-3)$. This point tells us that when oldsymbol{x=0}, the value of oldsymbol{y} must be oldsymbol{-3}. We can plug these values into our general equation and solve for oldsymbol{C}. This is how we find the particular solution.

Let's substitute oldsymbol{x=0} and oldsymbol{y=-3}: oldsymbol{-3 = \frac{1}{3} ((0)^2+1)^{\frac{3}{2}} + C}.

Simplify inside the parentheses: oldsymbol{-3 = \frac{1}{3} (0+1)^{\frac{3}{2}} + C}.

Which becomes: oldsymbol{-3 = \frac{1}{3} (1)^{\frac{3}{2}} + C}.

Since oldsymbol{1} raised to any power is still oldsymbol{1}, we have: oldsymbol{-3 = \frac{1}{3} (1) + C}.

So, oldsymbol{-3 = \frac{1}{3} + C}.

To find oldsymbol{C}, we just need to subtract oldsymbol{\frac{1}{3}} from both sides: oldsymbol{C = -3 - \frac{1}{3}}.

To perform this subtraction, let's get a common denominator. oldsymbol{-3} is the same as oldsymbol{-\frac{9}{3}}. So, oldsymbol{C = -\frac{9}{3} - \frac{1}{3}}.

This gives us oldsymbol{C = -\frac{10}{3}}.

Fantastic! We've found the value of our constant of integration. This is the final piece of the puzzle.

The Final Solution: Our Unique Function

We've successfully integrated the derivative and used the given point to find the constant of integration. Now, we put it all together to write down the specific function that satisfies both conditions. We simply take our general antiderivative oldsymbol{y = \frac{1}{3} (x^2+1)^{\frac{3}{2}} + C} and substitute the value of oldsymbol{C} we just found.

So, the final function is: oldsymbol{y = \frac{1}{3} (x^2+1)^{\frac{3}{2}} - \frac{10}{3}}.

And there you have it, folks! We've found the function whose derivative is oldsymbol{x \sqrt{x^2+1}} and passes through the point $(0,-3)$. It's a great feeling when all the steps click into place, right? This process of finding the original function from its derivative is fundamental in calculus, especially when dealing with problems involving rates of change, accumulation, and initial conditions. Keep practicing these u-substitution and point-solving techniques, and you'll be a calculus pro in no time! Remember, every complex problem can be broken down into simpler steps, just like how we turned a tricky integral into a power rule application. Keep exploring, keep questioning, and keep integrating!

Why This Matters: Real-World Connections

So, why is this whole process of finding the original function from its derivative and a point so darn important, guys? It's not just some abstract math exercise! This is the backbone of solving a ton of real-world problems. Think about physics, for instance. If you know the acceleration of an object (which is the second derivative of its position), and you know its initial velocity and initial position, you can use this exact method to find its position at any time. That's how we predict trajectories of rockets, analyze the motion of planets, or even design the suspension system for your car!

In economics, derivatives represent marginal costs or revenues. Finding the original function (the total cost or revenue function) from these marginal values, and knowing a specific value (like fixed costs), allows businesses to understand their overall financial picture, optimize pricing, and make informed decisions about production. Imagine trying to figure out the total profit of a company if you only knew how much profit each additional unit sold generated – you'd need to integrate!

Even in biology, population growth rates are often modeled using derivatives. If scientists can model the rate at which a population is changing, they can use integration to predict future population sizes, understand the impact of environmental factors, or plan conservation efforts. Knowing a specific population size at a certain time acts as that crucial point to anchor their predictions.

Essentially, whenever you have a situation described by a rate of change, and you have a known state at a particular moment, you can use the techniques we just practiced to reconstruct the whole picture over time or space. It's the difference between knowing how fast something is changing and knowing where it's actually going or what its total value is. So next time you're solving one of these problems, remember you're not just doing math homework; you're developing a powerful tool for understanding and predicting the world around you. Pretty cool, huh?

Common Pitfalls and How to Avoid Them

Okay, mathletes, let's talk about some of the common traps we can fall into when we're doing this kind of problem. Awareness is key to avoiding mistakes, right? The first big one is forgetting the constant of integration, C. Seriously, this trips up so many people! Remember, when you differentiate, any constant disappears. So, when you integrate, you must put that + C back in because you don't know what constant was there originally. If you forget it, you'll only find one possible function, not the entire family of functions that share that derivative. And then, when you try to use your point, you'll either get a nonsensical answer or realize something's wrong. Always, always remember the + C after the indefinite integral.

A close second is errors in u-substitution. This is where careful bookkeeping comes in. Make sure that when you substitute $u$, you also substitute the $dx$ part correctly. This means solving for $dx$ in terms of $du$ and the substituted expression. If your original integral had both $x$ and $u$ terms after your first substitution, you probably missed a step in converting everything to $u$. Double-check that oldsymbol{du = ( ext{derivative of } u) \, dx} step and make sure you can isolate the remaining parts of the integral.

Another common issue is arithmetic mistakes, especially when dealing with fractions and exponents. We saw this when we were solving for $C$. Adding or subtracting fractions incorrectly, or messing up with powers like oldsymbol{\frac{1}{2}+1 = \frac{3}{2}}, can completely derail your solution. Take your time with these calculations. Use a calculator if you need to, or work them out carefully on scratch paper. It's better to be a little slower and get it right than to rush and have to do it all over again.

Finally, some folks struggle with interpreting the given point. Remember that a point $(a, b)$ means that when the input variable (usually $x$) is $a$, the output variable (usually $y$) is $b$. So, you substitute $a$ for $x$ and $b$ for $y$ in your general equation. Don't mix them up or plug them into the wrong places. This is your anchor point, the specific piece of information that moves you from a general solution to a unique one.

By keeping these common pitfalls in mind – the missing oldsymbol{C}, u-substitution errors, arithmetic blunders, and misinterpreting points – you can navigate these integration problems much more smoothly. It's all about attention to detail and understanding the underlying logic. Keep practicing, and you'll start to anticipate these issues before they even happen!