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Differentiation graph question?
I have a function D= -0.4t^4 + 3.5t^3 - 0.5t^2 + 42t (distance travelled by a car for a 6 hour period, D=km and t=hours). I've found the derivative (-1.6t^3 + 10.5t^2 - t + 42), which I've graphed. However I'm stumped at how I use this answer to find when the car was speeding (i.e. over 80kph)? Is it the secant line formula? Thanks!
2 Answers
- 1 decade agoFavorite Answer
I think you need to understand the importance of the derivative and what it means. So you are given the distance function of a car. We all know the derivative is the rate of change of a function, but what does that really mean? Think of it this way, the derivative tells you the slope at a point.
So we have our distance function. Take time a and time b. The slope of the line between those two points is[ D(b) - D(a) ] / [ b - a ] another way of saying this is change in distance over change in time, or just distance over time. I hope you learned this in physics at some point! This is velocity! To be more precise it is the average velocity. Now the derivative is the same exact thing as this except it takes point b to be closer and closer to a, that the time between b and a is infintesimal. This gives us a slope at a single point, and in our case the instantaneous velocity at time t! Like a car's speedometer tells you the exact velocity of the car at that point in time.
So when you find the derivate of the distance function, you are getting the velocity function. It tells you the car's velocity at time t. So the problem is we need to find when the car is speeding, or over 80 kph. That is easy now that we know the velocity function! The question becomes for what values of t is -1.6t^3 + 10.5t^2 - t + 42) > 80 ? All you need to do is graph -1.6t^3 + 10.5t^2 - t + 42 and then graph 80. Find the interval when the function is above 80 and that is the time interval that the car was speeding.
I get when 2.4942 < t < 5.7301, the car is speeding.
I think I over explained alot, but I hope this helps you!
- cejaLv 45 years ago
discover any factors on each and every graph the place the tangent has gradient -a million. a million) y = 3 ? a million/3x³ dy/dx =?(a million/3)(3/x?) whilst the tangent has a gradient ?a million, it means that dy/dx= ?a million for this reason ?a million=?(a million/3)(3/x?) i.e. x? =a million i.e. x= a million for this reason y = 3 ? a million/{3×(a million)³}= 8/3 At (a million, 8/3), the tangent has a gradient ?a million 2) y= ??x dy/dx =?(½)/?x whilst the tangent has a gradient ?a million, it means that dy/dx= ?a million for this reason ?a million=?(½)/?x i.e. ?x =2 i.e. x= 4 for this reason y = ??4=?2 At is (4, ?2) , the tangent has a gradient ?a million