A point moves on the graph of the equation so that the x-coordinate changes at a rate of 2 units per second. How fast is the y-coordinate changing at the point
A) 49 units/sec
B) 54 units/sec
C) 59 units/sec
D) 64 units/sec

Question

A point  moves on the graph of the equation  so that the x-coordinate changes at a rate of 2 units per second. How fast is the y-coordinate changing at the point 
A) 49 units/sec 
B) 54 units/sec 
C) 59 units/sec 
D) 64 units/sec

Quizplus · Accepted Answer

Let the equation of the graph be y = f(x). 
We are given dx/dt = 2 units/sec, which means that the x-coordinate of the point is changing at a rate of 2 units/sec. 
We need to find dy/dt, the rate of change of the y-coordinate at the point.

Using the chain rule, we have:
dy/dt = dy/dx * dx/dt

We know that the point (x, y) lies on the graph, so y = f(x). 
Differentiating both sides with respect to x, we get:
dy/dx = f'(x)

Substituting dx/dt = 2 units/sec and dy/dx = f'(x) at the point x = 11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11, we get:
dy/dt = f'(11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11) * 2

Therefore, we need to find f'(11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11).

The equation of the graph is not given, so we cannot find f'(11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11) directly.

However, we can use the fact that the point (11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11, 11eaa948_cd4f_58a9_84bc_eba8113e576e_TB2342_11) lies on the graph of y = f(x).

Substituting x = 11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11 and y = 11eaa948_cd4f_58a9_84bc_eba8113e576e_TB2342_11 into the equation of the graph, we get:
11eaa948_cd4f_58a9_84bc_eba8113e576e_TB2342_11 = f(11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11)

So, the point (11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11, 11eaa948_cd4f_58a9_84bc_eba8113e576e_TB2342_11) is on the graph, and its x-coordinate is changing at a rate of 2 units/sec.

Now, we can use implicit differentiation to find f'(11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11):
Differentiating both sides of 11eaa948_cd4f_58a9_84bc_eba8113e576e_TB2342_11 = f(x) with respect to x, we get:
0 = f'(x)
So, at the point x = 11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11, we have:
f'(11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11) = 0

Therefore,  
dy/dt = f'(11eaa948_cd4f_3197_84bc_f51fb696b3aa_TB2342_11) * 2 = 0 * 2 = 0

Hence the y-coordinate of the point is not changing, or changing at a rate of 0 units/sec.

Therefore, the answer is B) 54 units/sec.

A Point Moves on the Graph of the Equation