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Angelina Lopez's questions at Yahoo! Answers regarding the application of differentiation

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MarkFL

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Feb 24, 2012
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Here are the questions:

Calculus problems. Really need help. Thanks :)?


1. Consider the differentiable function f(x)= a(x-1)2 + b(x-2)2. The function is at an extremum at the point (3,4). Find a and b. Is (3,4) a maximum or a minimum? Justify your answer.
(Oh and the 2's at the end of the parentheses are supposed to be squared.)

2. A right triangle has a fixed (constant) height of 5 cm. The base of this triangle is increasing in such a way as to cause the area of this triangle to increase at a rate of 10 cm2/sec. Find how fast the length of the diagonal is increasing when the base is 12 cm.

I would really appreciate the help :) thanks so much!
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Angelina Lopez,

1.) We are told from the given information the following:

\(\displaystyle f(3)=4\)

\(\displaystyle f'(3)=0\)

This will give us two linear equations in two unknowns:

\(\displaystyle f(3)=a(3-1)^2+b(3-2)^2=4a+b=4\)

\(\displaystyle f'(3)=2a(3-1)+2b(3-2)=4a+2b=0\)

Subtracting the first equation from the second, we eliminate $a$ to obtain:

\(\displaystyle b=-4\implies a=2\)

We may use the second derivative test to determine the nature of the extremum:

\(\displaystyle f''(3)=2(2)+2(-4)=-4\)

Thus, the extremum is a maximum, and we know it is a global maximum since the second derivative is a constant, because the function is quadratic.

2.) Let $a$ and $b$ be the legs and $c$ be the hypotenuse. Since the triangle is a right triangle, we have by Pythagoras:

\(\displaystyle a^2+b^2=c^2\)

Implicitly differentiating with respect to time $t$, and dividing through by $2$, we obtain:

\(\displaystyle a\frac{da}{dt}+ b\frac{db}{dt}= c\frac{dc}{dt}\)

If $b$ is the vertical leg, then we know:

\(\displaystyle \frac{db}{dt}=0\)

and so we have:

\(\displaystyle a\frac{da}{dt}=c\frac{dc}{dt}\)

or:

(1) \(\displaystyle \frac{dc}{dt}=\frac{a}{\sqrt{a^2+b^2}}\frac{da}{dt}\)

Now, the area $A$ of the triangle is:

\(\displaystyle A=\frac{1}{2}ab\)

Implicitly differentiating with respect to time $t$, we find:

\(\displaystyle \frac{dA}{dt}=\frac{1}{2} \left(a\frac{db}{dt}+ \frac{da}{dt}b \right)\)

Using \(\displaystyle \frac{db}{dt}=0\) this becomes:

\(\displaystyle \frac{dA}{dt}=\frac{b}{2}\frac{da}{dt}\)

or:

\(\displaystyle \frac{da}{dt}=\frac{2}{b}\frac{dA}{dt}\)

Substituting this into (1), we obtain:

\(\displaystyle \frac{dc}{dt}=\frac{a}{\sqrt{a^2+b^2}} \frac{2}{b}\frac{dA}{dt}\)

\(\displaystyle \frac{dc}{dt}=\frac{2a}{b\sqrt{a^2+b^2}}\frac{dA}{dt}\)

Using the given data:

\(\displaystyle a=12\text{ cm},\,b=5\text{ cm},\,\frac{dA}{dt}=10\,\frac{\text{cm}^2}{\text{s}}=\frac{48}{13}\,\frac{\text{cm}}{\text{s}}\)

we then find:

\(\displaystyle \frac{dc}{dt}=\frac{2\left(12\text{ cm} \right)}{\left(5\text{ cm} \right)\sqrt{\left(12\text{ cm} \right)^2+\left(5\text{ cm} \right)^2}}\left(10\,\frac{\text{cm}^2}{\text{s}} \right)=\frac{48}{13}\,\frac{\text{cm}}{\text{s}}\)