At what value of h is the rate of increase of √h twice the rate of increase of h?
(a) 1/16
(b) 1/4
(c) 1
(d) 2
(e) 4
answer is a = 1/16
Why did the mathematician go broke? Because he lost his fractions!
Now, onto the question. Let's find the rate of increase of √h and h first. The rate of increase of a function f is given by its derivative f'(h).
For √h, the derivative is 1/(2√h), and for h, the derivative is 1.
We want to find the value of h where the rate of increase of √h is twice the rate of increase of h. So, we can set up the equation:
1/(2√h) = 2(1)
Simplifying this equation, we get:
1/(2√h) = 2
Cross-multiplying, we have:
1 = 4√h
Squaring both sides, we get:
1 = 16h
Dividing both sides by 16, we find:
h = 1/16
Therefore, the value of h where the rate of increase of √h is twice the rate of increase of h is (a) 1/16.
Keep those math jokes coming!
To determine the value of h at which the rate of increase of √h is twice the rate of increase of h, we need to calculate the derivatives of both functions.
Let's start with the rate of increase of √h. The derivative of √h with respect to h can be found using the chain rule:
d/dh (√h) = (1/2) * (h^(-1/2))
Next, let's find the rate of increase of h. The derivative of h with respect to h is simply 1:
d/dh (h) = 1
Now, we can set up the equation and solve for the value of h:
(1/2) * (h^(-1/2)) = 2 * 1
Simplifying the equation:
(1/2) * (h^(-1/2)) = 2
Multiplying both sides by 2:
(h^(-1/2)) = 4
Taking the square of both sides to solve for h:
h^(-1/2) = 4^2
h^(-1/2) = 16
Taking the reciprocal of both sides:
h^(1/2) = 1/16
Squaring both sides:
h = (1/16)^2
Simplifying:
h = 1/256
Therefore, the answer is not among the options given.
To find the value of h at which the rate of increase of √h is twice the rate of increase of h, we need to differentiate both functions with respect to h and set up an equation.
Let's start by finding the derivatives of the functions √h and h:
1. √h:
To differentiate √h, we can use the power rule. Since √h can also be written as h^(1/2), the derivative is:
d(√h)/dh = (1/2) * h^(-1/2) = 1/(2√h)
2. h:
To differentiate h, we can use the power rule directly. The derivative is:
d(h)/dh = 1
Now, let's set up the equation based on the rates of increase:
The rate of increase of √h is given by the derivative d(√h)/dh, which is 1/(2√h).
The rate of increase of h is given by the derivative d(h)/dh, which is 1.
According to the problem, the rate of increase of √h is twice the rate of increase of h. Mathematically, we can write this as an equation:
1/(2√h) = 2 * 1
Simplifying the equation:
1/(2√h) = 2
Multiply both sides of the equation by 2:
1/√h = 4
Take the reciprocal of both sides of the equation:
√h = 1/4
Finding the square of both sides:
h = (1/4)^2
h = 1/16
Therefore, the value of h at which the rate of increase of √h is twice the rate of increase of h is 1/16.
So, the correct answer is (a) 1/16.
let y = h
dy/dh = 1
let y = √h = h^(1/2)
dy/dh = (1/2)h^(-1/2 = 1/(2√h)
so when is 1 = 2(1/2)/(2√h)
1 = 1/(2√h)
2√h = 1
√h=1/2
h = 1/4