Question 1192199: Variables x and y are such that, when e^y is plotted against x^2, a straight line graph passing through
the points (0.2, 1) and (0.5, 1.6) is obtained.
(i) Find the values of e^y when x = 0.
(ii) Express y in terms of x.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**(i) Find the values of e^y when x = 0:**
1. **Recognize the linear relationship:** Since plotting *eʸ* against *x²* gives a straight line, we can express this relationship as:
*eʸ = mx² + c*
where *m* is the slope and *c* is the y-intercept.
2. **Calculate the slope (m):** We are given two points on the line: (x², eʸ) = (0.2², 1) and (0.5², 1.6).
*m = (change in eʸ) / (change in x²)*
*m = (1.6 - 1) / (0.5² - 0.2²)*
*m = 0.6 / (0.25 - 0.04)*
*m = 0.6 / 0.21*
*m = 20/7*
3. **Calculate the y-intercept (c):** We can use either of the given points and the slope to find *c*. Let's use (0.2², 1):
*1 = (20/7)(0.2²) + c*
*1 = (20/7)(0.04) + c*
*1 = 0.8/7 + c*
*c = 1 - 0.8/7*
*c = 6.2/7*
4. **Find *eʸ* when *x = 0*:** Substitute *x = 0* into the equation *eʸ = mx² + c*:
*eʸ = (20/7)(0²) + 6.2/7*
*eʸ = 6.2/7*
Therefore, when *x = 0*, *eʸ = 6.2/7* (approximately 0.886).
**(ii) Express y in terms of x:**
1. **We have the equation:** *eʸ = mx² + c*
2. **Substitute the values of *m* and *c*:**
*eʸ = (20/7)x² + 6.2/7*
3. **Take the natural logarithm of both sides:**
*ln(eʸ) = ln((20/7)x² + 6.2/7)*
4. **Simplify:**
*y = ln((20/7)x² + 6.2/7)*
Therefore, *y = ln((20/7)x² + 6.2/7)*.
|
|
|
