MHB Angelina Lopez's Calculus Questions on Differentiation

AI Thread Summary
The discussion centers on two calculus problems involving differentiation. In the first problem, the function f(x) is analyzed to find constants a and b, concluding that a=2 and b=-4, indicating that the point (3,4) is a global maximum. The second problem involves a right triangle with a fixed height, where the area is increasing at a rate of 10 cm²/sec, leading to the calculation of the rate at which the diagonal is increasing when the base is 12 cm, resulting in a rate of 48/13 cm/sec. The solutions utilize implicit differentiation and the second derivative test for analysis. The thread provides a comprehensive approach to solving these calculus questions.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$\frac{dc}{dt}=\frac{2\left(12\text{ cm} \right)}{\left(5\text{ cm} \right)\sqrt{\left(12\text{ cm} \right)^2+\left(5\text{ cm} \right)^2}}\left(10\,\frac{\text{cm}^2}{\text{s}} \right)=\frac{48}{13}\,\frac{\text{cm}}{\text{s}}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
44
Views
4K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
3
Views
1K
Replies
1
Views
2K
Replies
9
Views
2K
Replies
1
Views
8K
Replies
2
Views
2K
Back
Top