Numerical approximations to definite integrals - AP Calculus AB

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Question

Write the equation of a tangent line to the given function at the point.

y = ln(x2) at (e, 3)

Answer

To solve this, first find the derivative of the function (otherwise known as the slope).

y = ln(x2)

y' = (2x/(x2))

Then, to find the slope in respect to the given points (e, 3), plug in e.

y' = (2e)/(e2)

Simplify.

y'=(2/e)

The question asks to find the tangent line to the function at (e, 3), so use the point-slope formula and the points (e, 3).

y – 3 = (2/e)(x – e)

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Question

If $f(x)=4, f'(x)=3, g(x)=2, \ and\ g'(x)=1$ then find .

$h(x)=\frac{2f}{g}$

Answer

The answer is 1.

$h(x)=\frac{2f}{g}$

$h'(x)=\frac{2(f'(x)g(x)-f(x)g'(x))}{g^{2}}$

$h'(x)=\frac{2(32-41)}{4} = 1$

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Question

If $f(x)=1, f'(x)=2, g(x)=3, \ and\ g'(x)=4$ then find .

Answer

The answer is 10.

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Question

Find the equation of the tangent line at on graph

Answer

The answer is

$f(x)=x^{2}+2x-8$

$f'(x)=2x+2$

(This is the slope. Now use the point-slope formula)

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Question

Find the equation of the tangent line at (1,1) in

$f(x)=ax^{2}+bx+c$

Answer

The answer is

$f(x)=ax^{2}+bx+c$

$f'(x)=2ax+b$

$f'(1)=2a(1)+b = 2a+b$

(This is the slope. Now use the point-slope formula.)

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Question

If $f(x)=\cos x$ then

$f^{'}\left ( \frac{\pi }{2} \right )=$

Answer

The answer is .

We know that

$\frac{\pi }{2}=90^{\circ}$

so,

$f'\left ( \frac{\pi }{2}\right )=-sin(90^{\circ})=-1$

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Question

Differentiate $y=3\csc x+5\sin x$

Answer

The answer is $-3\csc x\cot x+5\cos x$

We simply differentiate by parts, remembering our trig rules.

$y=3\csc x+5\sin x$

$y^{'}= -3\csc x\cot x+5\cos x$

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Question

Find the equation of the tangent line at when

$y=\frac{x^{3}-2x(x^{1/2})}{x}$

Answer

The answer is

$y=\frac{x^{3}-2x(x^{1/2})}{x}$ let's go ahead and cancel out the 's. This will simplify things.

$y=x^{2}-2(x^{1/2})$

$y'=2x-\frac{1}{(x^{1/2})}$

$y'(1)=2(1)-\frac{1}{(1^{1/2})} =1$ this is the slope so let's use the point slope formula.

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Question

Differentiate $y=\frac{t+2}{t^{2}-4t-12}$

Answer

We see the answer is $y'=\frac{-1}{(t-6)^{2}}$ after we simplify and use the quotient rule.

$y=\frac{t+2}{t^{2}-4t-12}$ we could use the quotient rule immediatly but it is easier if we simplify first.

$y=\frac{t+2}{(t+2)(t-6)}$

$y=\frac{1}{(t-6)}$

$y=(t-6)^{-1}$

$y'=-(t-6)^{-2}$

$y'=\frac{-1}{(t-6)^{2}}$

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Question

Find $\lim_{x\rightarrow \infty }\frac{-2x^3+x^2-2}{x^2+10}$

Answer

When taking limits to infinity, we usually only consider the highest exponents. In this case, the numerator has $-2x^3$ and the denominator has $x^2$ . Therefore, by cancellation, it becomes $-2x$ as approaches infinity. So the answer is $-\infty$ .

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Question

What is the first derivative of the function $\large h(x)=x^{\frac{1}{x}}$ ?

Answer

First, let .

$y=x^{\frac{1}{x}}$

We will take the natural logarithm of both sides in order to simplify the exponential expression on the right.

$\ln y=\ln x^{\frac{1}{x}}$

Next, apply the property of logarithms which states that, in general, $\log x^a=a\log x$ , where is a constant.

$\ln y = \frac{1}{x}\ln x$

We can differentiate both sides with respect to .

$\frac{d}{dx}\[ln y]=\frac{d}{dx}[\frac{1}{x}\ln x]$

We will need to apply the Chain Rule on the left side and the Product Rule on the right side.

$\frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{d}{dx}[\ln x]+\ln x\cdot \frac{d}{dx}[\frac{1}{x}]$

$\frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{1}{x} + \ln x\cdot \frac{-1}{x^{2}}$

Because we are looking for the derivative, we must solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$ .

$\frac{dy}{dx}=y\cdot \frac{1}{x^{2}}(1-\ln x)$

However, we want our answer to be in terms of only. We now substitute $x^{\frac{1}{x}}$ in place of .

$\frac{dy}{dx}=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

Since we let , we can replace $\frac{\mathrm{d} y}{\mathrm{d} x}$ with .

The answer is $h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$ .

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Question

Evaluate:

$\sum_{n=1}^{\infty }4(\frac{-1}{2})^{n-1}$

Answer

First, we can write out the first few terms of the sequence $a_{n}=$ $4(\frac{-1}{2})^{n-1}$ , where ranges from 1 to 3.

$a_{1}=4(\frac{-1}{2})^{1-1}=4$

$a_{2}=4(\frac{-1}{2})^{2-1}=-2$

$a_{3}=4(\frac{-1}{2})^{3-1}=1$

Notice that each term $\left ( n>1 \right )$ , is found by multiplying the previous term by $-\frac{1}{2}$ . Therefore, this sequence is a geometric sequence with a common ratio of $-\frac{1}{2}$ . We can find the sum of the terms in an infinite geometric sequence, provided that , where is the common ratio between the terms. Because $r=-\frac{1}{2}$ in this problem, $\left | r \right |$ is indeed less than 1. Therefore, we can use the following formula to find the sum, , of an infinite geometric series.

$S=\frac{a_{1}}{1-r}=\frac{4}{1-(\frac{-1}{2})}=\frac{4}{3/2}=\frac{8}{3}$

The answer is $\frac{8}{3}$ .

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Question

$\lim_{x\rightarrow 0}\frac{2\sin(x^2)}{x\tan(x)}=$

Answer

When we let x = 0 in our original limit, we obtain the 0/0 indeterminate form. Therefore, we can apply L'Hospital's Rule, which requires that we take the derivative of the numerator and denominator separately.

$\lim_{x\rightarrow 0}\frac{2\sin(x^2)}{x\tan(x)}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}2\sin(x^2)}{\frac{\mathrm{d} }{\mathrm{d} x}x\tan(x)}$

Apply the Chain Rule in the numerator and the Product Rule in the denominator.

$\lim_{x\rightarrow 0}\frac{4x\cos(x^2)}{\tan(x)+x\sec^2(x)}$

If we again substitute x = 0, we still obtain the 0/0 indeterminate form. Thus, we can apply L'Hospital's Rule one more time.

$\lim_{x\rightarrow 0}\frac{4x\cos(x^2)}{\tan(x)+x\sec^2(x)}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}4x\cos(x^2)}{\frac{\mathrm{d} }{\mathrm{d} x}(\tan(x)+x\sec^2(x))}$

$=\lim_{x\rightarrow 0}\frac{4\cos(x^2)+4x(2x)(-\sin(x^2))}{\sec^2(x)+(\sec^2(x)+x(2\sec^2x\tan x))}$

If we now let x = 0, we can evaluate the limit.

$=\lim_{x\rightarrow 0}\frac{4\cos(x^2)+4x(2x)(-\sin(x^2))}{\sec^2(x)+(\sec^2(x)+x(2\sec^2x\tan x))}=\frac{4\cos(0)+0}{\sec^2(0)+\sec^2(0)+0}$

$=\frac{4}{1+1}=2$

The answer is 2.

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Question

Consider the curve given by the parametric equations below:

$x(t)=\ln(t)$

What is the equation of the line normal to the curve when ?

Answer

In order to find the equation of the normal line, we will need the slope of the line and a point through which it passes. If we substitute into our parametric equations, we can easily obtain the point on the curve.

The normal line is perpendicular to the tangent line. Thus, we should first find the slope of the tangent line.

To find the value of the tangent slope when , we will use the following formula:

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}$

$y'(t)=3^t \ln 3 -2^t \ln 2$

$y'(1)=3\ln3-2\ln 2=\ln \frac{27}{4}$

$x'(t)=\frac{1}{t}$

$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\ln(\frac{27}{4})$

Because the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus,

slope of normal = $\frac{-1}{\ln(\frac{27}{4})}$ .

We now have the point and slope of the normal line, so we can use point-slope form.

$y-1=\frac{-1}{\ln \frac{27}{4}}(x-0)$

The answer is $y=\frac{-1}{\ln \frac{27}{4}}x+1$ .

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Question

Write the equation of a tangent line to the given function at the point.

y = ln(x2) at (e, 3)

Answer

To solve this, first find the derivative of the function (otherwise known as the slope).

y = ln(x2)

y' = (2x/(x2))

Then, to find the slope in respect to the given points (e, 3), plug in e.

y' = (2e)/(e2)

Simplify.

y'=(2/e)

The question asks to find the tangent line to the function at (e, 3), so use the point-slope formula and the points (e, 3).

y – 3 = (2/e)(x – e)

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Question

If $f(x)=4, f'(x)=3, g(x)=2, \ and\ g'(x)=1$ then find .

$h(x)=\frac{2f}{g}$

Answer

The answer is 1.

$h(x)=\frac{2f}{g}$

$h'(x)=\frac{2(f'(x)g(x)-f(x)g'(x))}{g^{2}}$

$h'(x)=\frac{2(32-41)}{4} = 1$

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Question

If $f(x)=1, f'(x)=2, g(x)=3, \ and\ g'(x)=4$ then find .

Answer

The answer is 10.

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Question

Find the equation of the tangent line at on graph

Answer

The answer is

$f(x)=x^{2}+2x-8$

$f'(x)=2x+2$

(This is the slope. Now use the point-slope formula)

Compare your answer with the correct one above

Question

Find the equation of the tangent line at (1,1) in

$f(x)=ax^{2}+bx+c$

Answer

The answer is

$f(x)=ax^{2}+bx+c$

$f'(x)=2ax+b$

$f'(1)=2a(1)+b = 2a+b$

(This is the slope. Now use the point-slope formula.)

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Question

If $f(x)=\cos x$ then

$f^{'}\left ( \frac{\pi }{2} \right )=$

Answer

The answer is .

We know that

$\frac{\pi }{2}=90^{\circ}$

so,

$f'\left ( \frac{\pi }{2}\right )=-sin(90^{\circ})=-1$

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