Question

A bug crawls along the graph of y=x²+4x+1. If its x-value is increasing at a rate of 3 cm/min, at what rate is its y-value increasing at the point (2,13) ?

Answer 1

To find the rate at which the y-value is increasing at the point (2, 13), we need to find the derivative of the function y = x² + 4x + 1 with respect to time and evaluate it at x = 2. The rate at which the x-value is changing is given as 3 cm/min. Plugging in the values, we find that the y-value is increasing at a rate of 24 cm/min.

Step-by-step explanation:

To find the rate at which the y-value is increasing, we need to find the derivative of the given function y = x² + 4x + 1 with respect to time.

Taking the derivative will give us the rate at which the y-value is changing with respect to the x-value, multiplied by the rate at which the x-value is changing with respect to time.

First, we find the derivative of the function: dy/dx = 2x + 4.

Next, we plug in the x-value of 2 into the derivative function: dy/dx = 2(2) + 4 = 8.

Finally, we multiply the rate at which the x-value is changing (3 cm/min) with the derivative at the point (2,13): (3 cm/min)(8) = 24 cm/min.