MHB Angelina Lopez's Calculus Questions on Differentiation

[MarkFL]

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.

 

[MarkFL]

Hello Angelina Lopez,

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

$$f(3)=4$$

$$f'(3)=0$$

This will give us two linear equations in two unknowns:

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

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

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

$$b=-4\implies a=2$$

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

$$f''(3)=2(2)+2(-4)=-4$$

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:

$$a^2+b^2=c^2$$

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

$$a\frac{da}{dt}+ b\frac{db}{dt}= c\frac{dc}{dt}$$

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

$$\frac{db}{dt}=0$$

and so we have:

$$a\frac{da}{dt}=c\frac{dc}{dt}$$

or:

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

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

$$A=\frac{1}{2}ab$$

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

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

Using $$\frac{db}{dt}=0$$ this becomes:

$$\frac{dA}{dt}=\frac{b}{2}\frac{da}{dt}$$

or:

$$\frac{da}{dt}=\frac{2}{b}\frac{dA}{dt}$$

Substituting this into (1), we obtain:

$$\frac{dc}{dt}=\frac{a}{\sqrt{a^2+b^2}} \frac{2}{b}\frac{dA}{dt}$$

$$\frac{dc}{dt}=\frac{2a}{b\sqrt{a^2+b^2}}\frac{dA}{dt}$$

Using the given data:

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

we then find:

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