Update README.md

README.md

@@ -9,6 +9,47 @@ tags:
 - medical
 datasets:
 - Locutusque/hyperion-v3.0
+widget:
+- example_title: Derivative
+  text: |
+    <|im_start|>user
+
+    In this task, you are given a description of an object and its surroundings. Your
+    task is to describe the motion of the object mentioned in the description using
+    differential equations. The answer should be a mathematical expression in t. Description:
+    A ball is thrown up at 30 ft/sec from 3 feet off the ground. Solve this step-by-step,
+    explaining each step in detail.<|im_end|>
+
+    <|im_start|>assistant
+  output:
+    text: |
+      Let's denote the height of the ball as h(t) where t represents time in seconds. Since the ball is thrown up with an initial velocity of 30 ft/sec, we can write the following equation for the vertical component of the ball's velocity:
+
+      v(t) = -32t + 30
+      
+      Here, v(t) represents the velocity of the ball at time t (in feet per second), and the negative sign indicates that the velocity decreases due to gravity. Now, we can integrate the velocity function to find the position function:
+      
+      h(t) = ∫v(t) dt = ∫(-32t + 30) dt
+      
+      Integrating each term separately, we get:
+      
+      h(t) = (-16t^2 + 30t + C)
+      
+      To determine the constant C, we use the initial condition that the ball is thrown from 3 feet above the ground. At t=0, the height is 3 feet:
+      
+      h(0) = 3
+      
+      Substituting t=0 into the equation for h(t):
+      
+      3 = (-16(0)^2 + 30(0) + C)
+      
+      Solving for C, we get:
+      
+      C = 3
+      
+      So the final equation for the height of the ball as a function of time is:
+      
+      h(t) = -16t^2 + 30t + 3
 ---
 # Locutusque/Hyperion-3.0-Yi-34B