A student is in a 20 week calc class. The function f(t)=50-50e^0.2t f(t) measures the students problem speed in t weeks
-how many problems per hour can a student solve at the beginning of the course
- how many problems per hour can a student solve halfway through the course
-at what rate is the students problem speed increasing 6 weeks into the course
at the beginning --- > t = 0
f(0) = 50 - 50e^0 = 0
half-way through course --- > t=10
f(10) = 50 - 50e^2 = -319
your equation makes no sense
Your function is a decreasing function.
Why would the student's speed decrease as the course progresses ?
Check for typo.
how to figure out rate
To find how many problems per hour a student can solve at the beginning of the course, we need to evaluate the function f(t) when t = 0. Plug in t = 0 into the given function:
f(0) = 50 - 50e^(0.2*0)
= 50 - 50e^0
= 50 - 50(1)
= 50 - 50
= 0
Therefore, at the beginning of the course, a student can solve 0 problems per hour.
To find how many problems per hour a student can solve halfway through the course, we need to evaluate the function f(t) when t = 10 (half of 20). Plug in t = 10 into the given function:
f(10) = 50 - 50e^(0.2*10)
= 50 - 50e^2
We can use a calculator to find the approximate value of e^2, which is approximately 7.389.
f(10) ≈ 50 - 50 * 7.389
≈ 50 - 369.45
≈ -319.45
Therefore, halfway through the course, a student can solve approximately -319.45 problems per hour (note: this means the student's problem-solving speed has decreased).
To find the rate at which the student's problem speed is increasing 6 weeks into the course, we need to find the derivative of the given function f(t) with respect to t. Differentiate the function f(t) using the chain rule:
f'(t) = -50 * 0.2 * e^(0.2t)
= -10e^(0.2t)
Now, plug in t = 6 into the derivative function f'(t):
f'(6) = -10e^(0.2*6)
= -10e^1.2
We can use a calculator to find the approximate value of e^1.2, which is approximately 3.320.
f'(6) ≈ -10 * 3.320
≈ -33.20
Therefore, 6 weeks into the course, the student's problem speed is increasing at a rate of approximately -33.20 problems per hour (note: the negative sign indicates a decrease in problem-solving speed).