# Thread: Calculus I Packet nearly complete!!! Just need help with these questions.

1. ## Calculus I Packet nearly complete!!! Just need help with these questions.

Hey everyone! I've been working on this huge packet filled with College Math II and Calculus 1 questions. I'm finished with 82 problems but now I'm stuck on the below few. If you can, please help me out because this is definitely URGENT!
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1. Given $y=2x^2-8$
Find the equation of the TANGENT line I at $x=2$.
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2. Given $y=[f(x)]^2 * [g(x)]^3$
Find $\frac{dy}{dx}$ "symbolically" (in prime notation)
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3. Given the position function $S=\frac{1}{4}t^4$, find the instantaneous acceleration at $t=4sec$
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4. $y=x\sqrt{2x+3}$
Find $\frac{dy}{dx}$
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5. Evaluate $\lim_{\Delta x\to 0} - \frac{4(x+\Delta x)-4y}{\Delta x}$
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6. The function $y = |x+21|$ is NOT differentiable at x=?
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7. Given the position $S(t)=2t+t^2$, where $S$ is in Ft.
Find the instantaneous velocity at $t=3sec$
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The questions above are the only ones I'm having a tough time with. Please help me! I would greatly appreciate it.

You guys are awesome! This site has been a great resource for me for the past 2 years. Keep it up!

4. $y=x\sqrt{2x+3}$
Find $\frac{dy}{dx}$
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6. The function $y = |x+21|$ is NOT differentiable at x=?
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I'll do these for you:

4.

$y = x\sqrt{2x+3}$

We simply have to use the product rule, which states:

If $f = x$ and $g = \sqrt{2x+3}$ and $y = fg$, then:

$y' = f'g + fg'$

So we just plug in:

$\frac{dy}{dx} = \sqrt{2x + 3} + \frac{2x}{2\sqrt{2x + 3}}$

Where we just used the chain rule for $\sqrt{2x + 3}$

Simplified:

$\frac{dy}{dx} = \sqrt{2x + 3} + \frac{x}{\sqrt{2x + 3}}$

Most teachers accept this answer, but if you need it any more simplified, here's how to go about it:

$\frac{dy}{dx} = \frac{2x + 3}{\sqrt{2x + 3}} + \frac{x}{\sqrt{2x + 3}}$

$\frac{dy}{dx} = \frac{3x + 3}{\sqrt{2x + 3}}$

And that's that one.

6. We have the function:

$y = |x + 21|$

We know that this is an absolute value function, and that at a single point it touches the x-axis and takes an immediate shot upwards to avoid negative values.

At this point is what is called a sharp point. Since it is a sharp point there, the derivative for the equation at that point doesn't exist.

That absolute value function can never be negative, so we can assume that the sharp point occurs at y = 0, so we just set the equation equal to 0:

$0 = |x + 21|$

Now we split the equation:

$0 = x + 21$
$0 = -x - 21$

Now we solve each one:

$x = -21$

$-x = 21$
$x = -21$

They both seem to give us the same value, and that makes it clear that both sides of the absolute value graph converge to this sharp point, meaning that this function is not differentiable at $x = -21$

3. Question 1:

$y -y_1 = m(x-x_1)$ will be the equation of the tangent.

where $(x_1 , y_1)$ is a point on the tangent.

use the derivative of the equation to find the value of m.

Question 2:

before you start know the chain and product rules ? If your familiar with them you just need to apply them very carefully.

Question 3:

Differentiate twice and, sub t = 4 .

Question 4:

Same as question 2, chain and product rule together.

Question 5:

hmmm, not sure, if it were $f(x+\Delta x)$ where $y = f(x)$ then it would be a lot easier

Question 6:

tried drawing a graph? the function has a sharp "turn" at one point, where the gradient just suddenly changes, that is the point you are looking for.

Question 7:

Differentiate once sub t = 3 and give the correct units.

Hey everyone! I've been working on this huge packet filled with College Math II and Calculus 1 questions. I'm finished with 82 problems but now I'm stuck on the below few. If you can, please help me out because this is definitely URGENT!
1. Given $y=2x^2-8$
Find the equation of the TANGENT line I at $x=2$.
Start with finding the tangent points coordinates. If $x=2, y=0$ so it is $(2,0)$.

Now, find the slope of the tangent at $x=2$. This slope is equal to $y'(2)$. So let's calculate it.
$y = 2x^2-8$
$y' = 4x$
$y'(2)=8$

We found that the tangent point is on the point $(2,0)$ and its slope is $8$.

$y = mx + C$
$y = 8x + C$
Plug $x=2$ and $y=0$ to find the constant C,
$0 = 16 + C$
$C = -16$

So the formula of the tangent line is:
$y=8x - 16$

2. Given $y=[f(x)]^2 * [g(x)]^3$
Find $\frac{dy}{dx}$ "symbolically" (in prime notation)
Take f(x) as f and g(x) as g. That'll cause no problem while we remember they're functions.

$y= f^2 g^3$
Differentiate both sides:
$y' = (f^2 g^3)'$
Due to the product rule,
$(f^2 g^3)' = (f^2)'(g^3) + (f^2)(g^3)'$
$(f^2)' = 2ff'$ and $(g^3)' = 3g^2g'$
Hence, $y' = 2fg^3f' + 3f^2g^2g'$

3. Given the position function $S=\frac{1}{4}t^4$, find the instantaneous acceleration at $t=4sec$
$S= \frac{1}{4}t^4$
If S is the position function, velocity is $\frac{dS}{dt}$ and the acceleration is $\frac{d^2S}{dt^2}$ which is the second derivative.

$S= \frac{1}{4}t^4$
$V = t^3$
$a = 3t^2$

So $a(4)=48$.

4. $y=x\sqrt{2x+3}$
Find $\frac{dy}{dx}$
This is an essential work. I leave this to you and recommend that you must solve it yourself. That's the only way you can learn. Hint: Product rule and Chain Rule..

5. Evaluate $\lim_{\Delta x\to 0} - \frac{4(x+\Delta x)-4y}{\Delta x}$
Is this the full question? If it is, the result is infinite, positive or negative depending on x and y. But I think you mistyped y for x.

6. The function $y = |x+21|$ is NOT differentiable at x=?
Hint: The graph is not differentiable at the corners. Function graphs containing absolute value has a corner. Where's it?

7. Given the position $S(t)=2t+t^2$, where $S$ is in Ft.
Find the instantaneous velocity at $t=3sec$
If the position function is $S=2t+t^2$, velocity is $V = \frac{dS}{dt}$. Find $V(3)$...

5. Thanks guys! I'm going to write down notes on what you all posted and study hard. I need to pass my quiz next week. :P

If anyone else wish to drop in a couple more notes, feel free! The more, the merrier I always say.